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  • Does Your Customer Engagement Create an Ah Feeling?

    - by Richard Lefebvre
    An (Oracle CX Blog) article by Christina McKeon Companies that successfully engage customers all have one thing in common. They make it seem easy for the customer to get what they need. No one would argue that brands don’t want to leave customers with this “ah” feeling. Since 94% of customers who have a low-effort service experience will buy from that company again, it makes financial sense for brands.1 Some brands are thinking differently about how they engage their customers to create ah feelings. How do they do it? Toyota is a great example of using smart assistance technology to understand customer intent and answer questions before customers hit the submit button online. What is unique in this situation is that Toyota captures intent while customers are filling out email forms. Toyota analyzes the data in the form and suggests responses before the customer sends the email. The customer gets the right answer, and the email never makes it to your contact center — which makes you and the customer happy. Most brands are fully aware of chat as a service channel, but some brands take chat to a whole new level. Beauty.com, part of the drugstore.com and Walgreens family of brands, uses live chat to replicate the personal experience that one would find at high-end department store cosmetic counters. Trained beauty advisors, all with esthetician or beauty counter experience, engage in live chat sessions with online shoppers to share immediate advice on the best products for their personal needs. Agents can watch customer activity online and determine the right time to reach out and offer help, just as help would be offered in a brick-and-mortar store. And, agents can co-browse along with the customer helping customers with online check-out. These personal chat discussions also give Beauty.com the opportunity to present products, advertise promotions, and resolve customer issues when they arise. Beauty.com converts approximately 25% of chat sessions into product orders. Photobox, the European market leader in online photo services, wanted to deliver personal and responsive service to its 24 million members. It ensures customer inquiries on personalized photo products are routed based on agent knowledge so customers get what they need from the company experts. By using a queuing system to ensure that the agent with the most appropriate knowledge handles the query, agent productivity increased while response times to 1,500 customer queries per day decreased. A real-time dashboard prevents agents from being overloaded with queries. This approach has produced financial results with a 15% increase in sales to existing customers and a 45% increase in orders from newly referred customers.

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  • Gartner PCC: A Shovel & Some Ah-Ha's

    - by kellsey.ruppel
    When Gartner Vice President and leading analyst Whit Andrews kicked off the Gartner Portals, Content & Collaboration Summit on Monday, March 12 at the Gaylord Palms in Orlando, FL by bringing a shovel to the stage, eyebrows raised and a few thoughts went through my head. Either this guy plans to go help the construction workers outside construct that new pool at the Gaylord or he took a wrong turn and is at the wrong conference. Oh and how did he get that shovel through airport security? As Whit explained more his objective became more clear…take everything anyone has ever told you about portals and throw it out the window, as portals have evolved and times they are most certainly changing. The future Web is here, available not only on browsers but also via a broad spectrum of access points, including automobiles, consumer electronics and more and more mobile devices. Not merely prevalent, the future Web is also multimedia-driven and operates in real time, driven by mobility, social media, streaming video and other dynamic services. Applications and user experiences are in the midst of an evolution — from the early, simple mobile Web models to today’s Web 2.0 mobile apps and, ultimately, to a world of predominantly Web apps. Additionally, cloud services will forever change how portals and user experience are designed, built, delivered, sourced and managed. So what does this mean for you? Today’s organizations need software that will enable them to not just do their jobs, but to do it in a way that is familiar and easy for them.  What does this mean for IT? Use software and technology as an enabler, not as a roadblock. Overall, we had a great week in Orlando learning about how to improve the user experience, manage content explosion, launch social initiatives, transition to mobile environments and understand cloud and SaaS options.  We had some great conversations throughout the conference and at the Oracle booth. Lots of demonstrations were given of Oracle WebCenter Sites and Oracle Social Network. And as Christie mentioned earlier this week, our Vice President of Product Management and Strategy for WebCenter Loren Weinberg presented on the topic of customer engagement and talked about how organization’s relationships with their customers have fundamentally changed today and the resulting impact that has on their priorities.  Loren also talked about the importance of customer engagement, why that matters now more than ever, and what you can do to help your company or organization succeed in this new world. The question asked in every keynote and session was a simple one: What is your “ah-ha” moment? I personally had quite a few, some of which I’ve captured below. 70% of internal social initiatives eventually fail. By 2014, refusing to communicate with consumers via social media will be as harmful as ignoring emails/phone calls is today. Customer engagement = multi-channel + social & interactive + personal & relevant + optimized. If people choose to talk about your product/company/service, it's because it's remarkable. -- Seth Godin's keynote (one of the highlights of the conference!) The Web will become the primary method used for delivering content and applications to mobile devices. By 2015, 20% of smart phone users worldwide will conduct commerce using context-enriched services on a weekly basis.  86% of customers will pay more for a better customer experience. 6 P's of Quality User Experience. Product. Enabled by: People, Patterns, Process, Profit, Priorities. Did you attend the Gartner Summit? What were your ah-ha moments?

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  • My "Ah-Ha!" Moment With LINQ

    - by CompiledMonkey
    I'm currently working on a set of web services that will be consumed by iPhone and Android devices. Given how often the web services will be called in a relatively short period of time, the data access for the web services has proven to be a very important aspect of the project. In choosing the technology stack for implementation, I opted for LINQ to SQL as it was something I had dabbled with in the past and wanted to learn more about in a real environment. The query optimization happening behind...(read more)

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  • Interrupt ?13 (ah=48) - don't working

    - by GLeBaTi
    mov dl,00h mov ah,08h int 13h this is code showing normal parameters of floppy disk. mov dl,80h mov ah,08h int 13h this is code, showing not valid parameters of hard disk(may be, my hard disk space is big (LBA)), And I've written this code: mov dl,80h mov ah,48h int 13h it is code doing cf = 1(error). How fix it? I want learn parameters of my hard disk. (http://lrs.uni-passau.de/support/doc/interrupt-57/RB-0677.HTM)

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  • Interrupt ?13 (ah=48) - not working

    - by GLeBaTi
    I want fetch the parameters of my hard disk. Using the technique described here. This is code showing normal parameters of floppy disk: mov dl,00h mov ah,08h int 13h This is code, showing not valid parameters of hard disk (may be, my hard disk space is big (LBA)): mov dl,80h mov ah,08h int 13h And I've written this code: mov dl,80h mov ah,48h int 13h The code is giving cf = 1(error). How do I fix it?

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  • Python - multithreading / multiprocessing, very strange problem.

    - by orokusaki
    import uuid import time import multiprocessing def sleep_then_write(content): time.sleep(5) print(content) if __name__ == '__main__': for i in range(15): p = multiprocessing.Process(target=sleep_then_write, args=('Hello World',)) p.start() print('Ah, what a hard day of threading...') This script output the following: Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... AAh, what a hard day of threading.. h, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Ah, what a hard day of threading... Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Hello World Firstly, why the heck did it print the bottom statement sixteen times (one for each process) instead of just the one time? Second, notice the AAh, and h, about half way down; that was the real output. This makes me wary of using threads ever, now. (Windows XP, Python 2.6.4, Core 2 Duo)

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  • 8086 programming using TASM: pc to pc communication

    - by Komal
    .model small .stack 100 .data .code mov ah,00h mov al,0e3h mov dx,00h int 14h back: nop l1: mov ah,03h mov dx,00h int 14h and ah,01h cmp ah,01h jne l1 mov ah,02h mov dx,00h int 21h mov dl,al mov ah,02h int 21h jmb back mov ah,4ch int 21h end this a pc to pc commnication receiver program.i would like to know why have we used the mov dx,00h function and what is the meaning of mov al,0e3h this ?

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  • fail2ban custom action to permanent ban IPs from China

    - by John Magnolia
    When a IP address gets banned how can I check if the banned IP address is from China. If yes, then add it to the permanent ban list. I have found this nice guide which write the banned IP to file. Reason: I am getting a lot of brute force attacks from China daily, thankfully fail2ban is helping restrict this although they appear to be getting worse and they are just changing their IP Address. Or even better would be if there was a maintained database of known hacker IP addresses. Example 1 Hi, The IP 60.169.78.77 has just been banned by Fail2Ban after 4 attempts against vsftpd. Here are more information about 60.169.78.77: % [whois.apnic.net node-7] % Whois data copyright terms http://www.apnic.net/db/dbcopyright.html inetnum: 60.166.0.0 - 60.175.255.255 netname: CHINANET-AH descr: CHINANET anhui province network descr: China Telecom descr: A12,Xin-Jie-Kou-Wai Street descr: Beijing 100088 country: CN admin-c: CH93-AP tech-c: JW89-AP mnt-by: APNIC-HM mnt-routes: MAINT-CHINANET-AH mnt-lower: MAINT-CHINANET-AH status: ALLOCATED PORTABLE changed: [email protected] 20040721 source: APNIC person: Chinanet Hostmaster nic-hdl: CH93-AP e-mail: [email protected] address: No.31 ,jingrong street,beijing address: 100032 phone: +86-10-58501724 fax-no: +86-10-58501724 country: CN changed: [email protected] 20070416 mnt-by: MAINT-CHINANET source: APNIC person: Jinneng Wang address: 17/F, Postal Building No.120 Changjiang address: Middle Road, Hefei, Anhui, China country: CN phone: +86-551-2659073 fax-no: +86-551-2659287 e-mail: [email protected].ah.cninfo.net nic-hdl: JW89-AP mnt-by: MAINT-NEW changed: [email protected].ah.cninfo.net 19990818 source: APNIC Regards, Fail2Ban Example 2 Hi, The IP 60.169.78.81 has just been banned by Fail2Ban after 4 attempts against vsftpd. Here are more information about 60.169.78.81: % [whois.apnic.net node-6] % Whois data copyright terms http://www.apnic.net/db/dbcopyright.html inetnum: 60.166.0.0 - 60.175.255.255 netname: CHINANET-AH descr: CHINANET anhui province network descr: China Telecom descr: A12,Xin-Jie-Kou-Wai Street descr: Beijing 100088 country: CN admin-c: CH93-AP tech-c: JW89-AP mnt-by: APNIC-HM mnt-routes: MAINT-CHINANET-AH mnt-lower: MAINT-CHINANET-AH status: ALLOCATED PORTABLE changed: [email protected] 20040721 source: APNIC person: Chinanet Hostmaster nic-hdl: CH93-AP e-mail: [email protected] address: No.31 ,jingrong street,beijing address: 100032 phone: +86-10-58501724 fax-no: +86-10-58501724 country: CN changed: [email protected] 20070416 mnt-by: MAINT-CHINANET source: APNIC person: Jinneng Wang address: 17/F, Postal Building No.120 Changjiang address: Middle Road, Hefei, Anhui, China country: CN phone: +86-551-2659073 fax-no: +86-551-2659287 e-mail: [email protected].ah.cninfo.net nic-hdl: JW89-AP mnt-by: MAINT-NEW changed: [email protected].ah.cninfo.net 19990818 source: APNIC Regards, Fail2Ban Example 3 Hi, The IP 222.133.244.99 has just been banned by Fail2Ban after 4 attempts against vsftpd. Here are more information about 222.133.244.99: % [whois.apnic.net node-6] % Whois data copyright terms http://www.apnic.net/db/dbcopyright.html inetnum: 222.133.244.96 - 222.133.244.127 netname: LCZFFHQ country: CN descr: liaochenggovermentfanghuoqiang admin-c: DS95-AP tech-c: DS95-AP status: ASSIGNED NON-PORTABLE changed: [email protected] 20060122 mnt-by: MAINT-CNCGROUP-SD source: APNIC route: 222.132.0.0/14 descr: CNC Group CHINA169 Shandong Province Network country: CN origin: AS4837 mnt-by: MAINT-CNCGROUP-RR changed: [email protected] 20060118 source: APNIC person: Data Communication Bureau Shandong nic-hdl: DS95-AP e-mail: [email protected] address: No.77 Jingsan Road,Jinan,Shandong,P.R.China phone: +86-531-6052611 fax-no: +86-531-6052414 country: CN changed: [email protected] 20050330 mnt-by: MAINT-CNCGROUP-SD source: APNIC Regards, Fail2Ban

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  • Runtime error of TASM language help!

    - by dominoos
    .model small .stack 400h .data message db "hello. ", 0ah, 0dh, "$" firstdigit db ? seconddigit db ? thirddigit db ? number dw ? newnumber db ? anumber dw 0d bnumber dw 0d Firstn db 0ah, 0dh, "Enter first 3 digit number: ","$" secondn db 0ah, 0dh, "Enter second 3 digit number: ","$" messageB db 0ah, 0dh, "HCF of two number is: ","$" linebreaker db 0ah, 0dh, ' ', 0ah, 0dh, '$' .code Start: mov ax, @data ; establish access to the data segment mov ds, ax ; mov number, 0d mov dx, offset message ; print the string "yob choi 0648293" mov ah, 9h int 21h num: mov dx, offset Firstn ; print the string "put 1st 3 digit" mov ah, 9h int 21h ;run JMP FirstFirst ; jump to FirstFirst FirstFirst: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add anumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add anumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add anumber, ax ; jmp num2 ;go to checks Num2: mov dx, offset secondn ; print the string "put 2nd 3 digits" mov ah, 9h int 21h ;run JMP SecondSecond SecondSecond: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add bnumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add bnumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add bnumber, ax ; jmp compare ;go to compare compare: CMP ax, anumber ;comparing numbB and Number JA comp1 ;go to comp1 if anumber is bigger CMP ax, anumber ; JB comp2 ;go to comp2 if anumber is smaller CMP ax, anumber ; JE equal ;go to equal if two numbers are the same JMP compare ;go to compare (avioding error) comp1: SUB ax, anumber; subtract smaller number from bigger number JMP compare ; comp2: SUB anumber, ax; subtract smaller number from bigger number JMP compare ; equal: mov ah, 9d ;make linkbreak after the 2nd 3 digit number mov dx, offset linebreaker int 21h mov ah, 9d ;print "HCF of two number is:" mov dx, offset messageB int 21h mov ax,anumber ;copying 2nd number into ax add al,30h ; converting to ascii mov newnumber,al ; copying from low part of register into newnumb mov ah, 2d ;bios code for print a character mov dl, newnumber ;we had saved the ascii code here int 21h ;call to bios JMP exit; exit: mov ah, 4ch int 21h ;exit the program End hi, this is a program that finds highest common factor of 2 different 3digit number. if i put 200, 235,312 (low numbers) it works fine. but if i put 500, 550, 654(bigger number) the program crashes after the 2nd 3digit number is entered. can you help me to find out what problem is?

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  • adresse book with C programming, i have problem with library i think, couldn't complite my code

    - by osabri
    I've divided my code in small programm so it can be easy to excute /* ab_error.c : in case of errors following messages will be displayed */ #include "adressbook.h" static char *errormsg[] = { "", "\nNot enough space on disk", "\nCannot open file", "\nCannot read file", "\nCannot write file" }; void check(int error) { switch(error) { case 0: return; case 1: write_file(); case 2: case 3: case 4: system("cls"); fputs(errormsg[error], stderr); exit(error); } } 2nd /* ab_fileio.c : functions for file input/output */ include "adressbook.h" static char ab_file[] = "ADRESSBOOK.DAT"; //file to save the entries int read_file(void) { int error = 0; FILE *fp; ELEMENT *new_e, *last_e = NULL; DATA buffer; if( (fp = fopen(ab_file, "rb")) == NULL) return -1; //no file found while (fread(&buffer, sizeof(DATA), 1, fp) == 1) //reads one list element after another { if( (new_e = make_element()) == NULL) { error = 1; break; //not enough space } new_e->person = buffer; //copy data to new element new_e->next = NULL; if(hol.first == NULL) //list is empty? hol.first = new_e; //yes else last_e->next = new_e; //no last_e = new_e; ++hol.amount; } if( !error && !feof(fp) ) error = 3; //cannot read file fclose(fp); return error; } /-------------------------------/ int write_file(void) { int error = 0; FILE *fp; ELEMENT *p; if( (p = hol.first) == NULL) return 0; //list is empty if( (fp = fopen(ab_file, "wb")) == NULL) return 2; //cannot open while( p!= NULL) { if( fwrite(&p->person, sizeof(DATA), 1, fp) < 1) { error = 4; break; //cannot write } p = p->next; } fclose(fp); return error; } 3rd /* ab_list.c : functions to manipulate the list */ #include "adressbook.h" HOL hol = {0, NULL}; //global definition for head of list /* -------------------- */ ELEMENT *make_element(void) { return (ELEMENT *)malloc( sizeof(ELEMENT) ); } /* -------------------- */ int ins_element( DATA *newdata) { ELEMENT *new_e, *pre_p; if((new_e = make_element()) == NULL) return 1; new_e ->person = *newdata; // copy data to new element pre_p = search(new_e->person.family_name); if(pre_p == NULL) //no person in list { new_e->next = hol.first; //put it to the begin hol.first = new_e; } else { new_e->next = pre_p->next; pre_p->next = new_e; } ++hol.amount; return 0; } int erase_element( char name, char surname ) { return 0; } /* ---------------------*/ ELEMENT *search(char *name) { ELEMENT *sp, *retp; //searchpointer, returnpointer retp = NULL; sp = hol.first; while(sp != NULL && sp->person.family_name != name) { retp = sp; sp = sp->next; } return(retp); } 4th /* ab_screen.c : functions for printing information on screen */ #include "adressbook.h" #include <conio.h> #include <ctype.h> /* standard prompts for in- and output */ static char pgmname[] = "---- Oussama's Adressbook made in splendid C ----"; static char options[] = "\ 1: Enter new adress\n\n\ 2: Delete entry\n\n\ 3: Change entry\n\n\ 4: Print adress\n\n\ Esc: Exit\n\n\n\ Your choice . . .: "; static char prompt[] = "\ Name . . . .:\n\ Surname . . :\n\n\ Street . . .:\n\n\ House number:\n\n\ Postal code :\n\n\ Phone number:"; static char buttons[] = "\ <Esc> = cancel input <Backspace> = correct input\ <Return> = assume"; static char headline[] = "\ Name Surname Street House Postal code Phone number \n\ ------------------------------------------------------------------------"; static char further[] = "\ -------- continue with any key --------"; /* ---------------------------------- */ int menu(void) //show menu and read user input { int c; system ("cls"); set_cur(0,20); puts(pgmname); set_cur(6,0); printf("%s", options); while( (c = getch()) != ESC && (c < '1' || c > '4')) putch('\a'); return c; } /* ---------------------------------- */ int print_adr_book(void) //display adressbook { int line = 1; ELEMENT *p = hol.first; system("cls"); set_cur(0,20); puts(pgmname); set_cur(2,0); puts(headline); set_cur(5,0); while(p != NULL) //run through list and show entries { printf("%5d %-15s ",line, p->person.family_name); printf("%-12s %-15s ", p->person.given_name, p->person.street); printf("%-4d %-5d %-12d\n",p->person.house_number, p->person.postal_code, p->person.phone); p = p->next; if( p == NULL || ++line %16 == 1) //end of list or screen is full { set_cur(24,0); printf("%s",further); if( getch() == ESC) return 0; set_cur(5,0); scroll_up(0,5,24);//puts(headline); } } return 0; } /* -------------------------------------------*/ int make_entry(void) { char cache[50]; DATA newperson; ELEMENT *p; while(1) { system("cls"); set_cur(0,20); puts(pgmname); set_cur(6,0); puts("Please enter new data:"); set_cur(10,0); puts(prompt); set_cur(24,0); printf("%s",buttons); balken(10, 25, MAXL, ' ',0x70); //input name if(input(newperson.family_name, MAXL, ESC, CR) == ESC) return 0; balken(12,25, MAXL, ' ', 0x70); //surname if(input(newperson.given_name, MAXL, ESC, CR) == ESC) return 0; balken(14,25, 30, ' ', 0x70); //street if(input(newperson.street, 30, ESC, CR) == ESC) return 0; balken(16,25, 4, ' ',0x70); //housenumber if(input(cache, 4, ESC, CR) == ESC) return 0; newperson.house_number = atol(cache); //to string balken(18,25, 5, ' ',0x70); //postal code if(input(cache, 5, ESC, CR) == ESC) return 0; newperson.postal_code = atol(cache); //to string balken(20,25, 20, ' ',0x70); //phone number if(input(cache, 20, ESC, CR) == ESC) return 0; newperson.phone = atol(cache); //to string p = search(newperson.phone); if( p!= NULL && p->person.phone == newperson.phone) { set_cur(22,25); puts("phonenumber already exists!"); set_cur(24,0); printf("%s, further"); getch(); continue; } } } 5th /* adress_book_project.c : main program to create an adressbook */ /* copyrights by Oussama Sabri, June 2010 */ #include "adressbook.h" //project header file int main() { int rv, cmd; //return value, user command if ( (rv = read_file() ) == -1) // no data saved yet rv = make_entry(); check(rv); //prompts an error and quits program on disfunction do { switch (cmd = menu())//calls menu and gets user input back { case '1': rv = make_entry(); break; case '2': //delete entry case '3': //changes entry rv = change_entry(cmd); break; case '4': //prints adressbook on screen rv = print_adr_book(); break; case ESC: //end of program system ("cls"); rv = 0; break; } }while(cmd!= ESC); check ( write_file() ); //save adressbook return 0; } 6th /* Getcb.c --> Die Funktion getcb() liefert die naechste * * Tastatureingabe (ruft den BIOS-INT 0x16 auf). * * Return-Wert: * * ASCII-Code bzw. erweiterter Code + 256 */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> int getcb(void) { union REGS intregs; intregs.h.ah = 0; // Subfunktion 0: ein Zeichen // von der Tastatur lesen. int86( 0x16, &intregs, &intregs); if( intregs.h.al != 0) // Falls ASCII-Zeichen, return (intregs.h.al); // dieses zurueckgeben. else // Sonst den erweiterten return (intregs.h.ah + 0x100); // Code + 256 } 7th /* PUTCB.C --> enthaelt die Funktionen * * - putcb() * * - putcb9() * * - balken() * * - input() * * * * Es werden die Funktionen 9 und 14 des Video-Interrupts * * (ROM-BIOS-Interrupt 0x10) verwendet. * * * * Die Prototypen dieser Funktionen stehen in BIO.H */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> #define VIDEO_INT 0x10 /*---------------------------------------------------------------- * putcb(c) gibt das Zeichen auf der aktuellen Cursor-Position * am Bildschirm aus. Der Cursor wird versetzt. * Steuerzeichen Back-Space, CR, LF und BELL werden * ausgefuehrt. * Return-Wert: keiner */ void putcb(unsigned char c) /* Gibt das Zeichen in c auf */ { /* den Bildschirm aus. */ union REGS intregs; intregs.h.ah = 14; /* Subfunktion 14 ("Teletype") */ intregs.h.al = c; intregs.h.bl = 0xf; /* Vordergrund-Farbe im */ /* Grafik-Modus. */ int86(VIDEO_INT, &intregs, &intregs); } /*---------------------------------------------------------------- * putcb9(c,count,mode) gibt das Zeichen in c count-mal im * angegebenen Modus auf der aktuellen * Cursor-Position am Bildschirm aus. * Der Cursor wird nicht versetzt. * * Return-Wert: keiner */ void putcb9( unsigned char c, /* das Zeichen */ unsigned count, /* die Anzahl */ unsigned mode ) /* Low-Byte: das Atrribut */ { /* High-Byte: die Bildschirmseite*/ union REGS intregs; intregs.h.ah = 9; /* Subfunktion 9 des Int 0x10 */ intregs.h.al = c; intregs.x.bx = mode; intregs.x.cx = count; int86( VIDEO_INT, &intregs, &intregs); } /*---------------------------------------------------------------- * balken() positioniert den Cursor und zeichnet einen Balken, * wobei Position, L„nge, Fllzeichen und Attribut * als Argumente bergeben werden. * Der Cursor bleibt auf der ersten Position im Balken. */ void balken( unsigned int zeile, /* Start-Position */ unsigned int spalte, unsigned int laenge, /* Laenge des Balkens */ unsigned char c, /* Fuellzeichen */ unsigned int modus) /* Low-Byte: Attribut */ /* High-Byte: Bildschirmseite */ { union REGS intregs; intregs.h.ah = 2; /* Cursor auf der angegebenen */ intregs.h.dh = zeile; /* Bildschirmseite versetzen. */ intregs.h.dl = spalte; intregs.h.bh = (modus >> 8); int86(VIDEO_INT, &intregs, &intregs); putcb9(c, laenge, modus); /* Balken ausgeben. */ } /*---------------------------------------------------------------- * input() liest Zeichen von der Tastatur ein und haengt '\0' an. * Mit Backspace kann die Eingabe geloescht werden. * Das Attribut am Bildschirm bleibt erhalten. * * Argumente: 1. Zeiger auf den Eingabepuffer. * 2. Anzahl maximal einzulesender Zeichen. * 3. Die optionalen Argumente: Zeichen, mit denen die * Eingabe abgebrochen werden kann. * Diese Liste muá mit CR = '\r' enden! * Return-Wert: Das Zeichen, mit dem die Eingabe abgebrochen wurde. */ #include <stdarg.h> int getcb( void); /* Zum Lesen der Tastatur */ int input(char *puffer, int max,... ) { int c; /* aktuelles Zeichen */ int breakc; /* Abruchzeichen */ int nc = 0; /* Anzahl eingelesener Zeichen */ va_list argp; /* Zeiger auf die weiteren Arumente */ while(1) { *puffer = '\0'; va_start(argp, max); /* argp initialisieren */ c = getcb(); do /* Mit Zeichen der Abbruchliste vergleichen */ if(c == (breakc = va_arg(argp,int)) ) return(breakc); while( breakc != '\r' ); va_end( argp); if( c == '\b' && nc > 0) /* Backspace? */ { --nc; --puffer; putcb(c); putcb(' '); putcb(c); } else if( c >= 32 && c <= 255 && nc < max ) { ++nc; *puffer++ = c; putcb(c); } else if( nc == max) putcb('\7'); /* Ton ausgeben */ } } 8th /* Video.c --> Enthaelt die Funktionen * cls(), * scroll_up(), scroll_down(), * set_cur(), get_cur(), * set_screen_page(), get_screen_page() * * Die Prototypen dieser Funktionen befinden sich in BIO.H */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> #include "bio.h" #define VIDEO_INT 0x10 typedef unsigned char BYTE; void scroll_up( int anzahl, int anf_zeile, int end_zeile) { /* Fenster hoch rollen. */ union REGS intregs; intregs.x.ax = 0x600 + anzahl; /* Subfunktion AH = 6, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void scroll_down( int anzahl, int anf_zeile, int end_zeile) { /* Fenster runter rollen. */ union REGS intregs; intregs.x.ax = 0x700 + anzahl; /* Subfunktion AH = 7, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void set_cur( int zeile, int spalte) /* versetzt den Cursor */ { /* der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 2; intregs.h.dh = (BYTE)zeile; intregs.h.dl = (BYTE)spalte; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); } void get_cur(int *zeile, int *spalte) /* holt die Cursor- */ { /* Position der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 3; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); *zeile = (unsigned)intregs.h.dh; *spalte = (unsigned)intregs.h.dl; } void cls(void) { scroll_up(0,0,24); /* Gesamten Bildschirm loeschen. */ set_cur(0,0); /* Cursor in Home-Position. */ } int get_screen_page(void) /* Aktuelle Bildschirmseite holen.*/ { union REGS intregs; intregs.h.ah = 15; /* Subfunktion AH = 15: */ /* Bildschirm-Modus feststellen. */ int86(VIDEO_INT, &intregs, &intregs); return (intregs.h.bh); } void set_screen_page(int seite) /* setzt die aktive Seite des */ { /* Bildschirmpuffers auf die */ /* angegebene Seite. */ union REGS intregs; intregs.x.ax = 0x500 + seite; /* Subfunktion AH = 5 */ int86(VIDEO_INT, &intregs, &intregs); } /* ------------------------------------------------------------- Ein kleines Testprogramm : */ /* #include <stdio.h> int main() { cls(); set_cur(23, 0); printf("Weiter mit <Return>\n"); set_cur(12, 20); printf("Ein Test!\n"); getchar(); scroll_up(3, 5, 20); getchar(); scroll_down(6, 5, 20); getchar(); set_screen_page(1); printf("\nAuf der 2. Seite !\n"); getchar(); set_screen_page(0); set_cur(0,0); printf("\nWieder auf der 1. Seite !\n"); getchar(); cls(); return 0; } */ /* Video.c --> Enthaelt die Funktionen * cls(), * scroll_up(), scroll_down(), * set_cur(), get_cur(), * set_screen_page(), get_screen_page() * * Die Prototypen dieser Funktionen befinden sich in BIO.H */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> #include "bio.h" #define VIDEO_INT 0x10 typedef unsigned char BYTE; void scroll_up( int anzahl, int anf_zeile, int end_zeile) { /* Fenster hoch rollen. */ union REGS intregs; intregs.x.ax = 0x600 + anzahl; /* Subfunktion AH = 6, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void scroll_down( int anzahl, int anf_zeile, int end_zeile) { /* Fenster runter rollen. */ union REGS intregs; intregs.x.ax = 0x700 + anzahl; /* Subfunktion AH = 7, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void set_cur( int zeile, int spalte) /* versetzt den Cursor */ { /* der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 2; intregs.h.dh = (BYTE)zeile; intregs.h.dl = (BYTE)spalte; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); } void get_cur(int *zeile, int *spalte) /* holt die Cursor- */ { /* Position der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 3; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); *zeile = (unsigned)intregs.h.dh; *spalte = (unsigned)intregs.h.dl; } void cls(void) { scroll_up(0,0,24); /* Gesamten Bildschirm loeschen. */ set_cur(0,0); /* Cursor in Home-Position. */ } int get_screen_page(void) /* Aktuelle Bildschirmseite holen.*/ { union REGS intregs; intregs.h.ah = 15; /* Subfunktion AH = 15: */ /* Bildschirm-Modus feststellen. */ int86(VIDEO_INT, &intregs, &intregs); return (intregs.h.bh); } void set_screen_page(int seite) /* setzt die aktive Seite des */ { /* Bildschirmpuffers auf die */ /* angegebene Seite. */ union REGS intregs; intregs.x.ax = 0x500 + seite; /* Subfunktion AH = 5 */ int86(VIDEO_INT, &intregs, &intregs); } /* ------------------------------------------------------------- Ein kleines Testprogramm : */ /* #include <stdio.h> int main() { cls(); set_cur(23, 0); printf("Weiter mit <Return>\n"); set_cur(12, 20); printf("Ein Test!\n"); getchar(); scroll_up(3, 5, 20); getchar(); scroll_down(6, 5, 20); getchar(); set_screen_page(1); printf("\nAuf der 2. Seite !\n"); getchar(); set_screen_page(0); set_cur(0,0); printf("\nWieder auf der 1. Seite !\n"); getchar(); cls(); return 0; } */ /* BIO.H --> Enthaelt die Prototypen der BIOS-Funktionen. */ /* --- Funktionen in VIDEO.C --- */ extern void scroll_up(int anzahl, int anf_zeile,int end_zeile); extern void scroll_down(int anzahl, int anf_zeile, int end_zeile); extern void set_cur(int zeile, int spalte); extern void get_cur(int *zeile, int *spalte); extern void cls(void); extern int get_screen_page(void); extern void set_screen_page(int page); /* --- Funktionen in GETCB.C / PUTCB.C --- */ extern int getcb(void); extern void putcb(int c); extern void putcb9(int c, unsigned count, unsigned modus); extern void balken(int zeile, int spalte, int laenge, int c, unsigned modus); extern int input(char *puffer, int max,... ); need your help, can't find my mistakes:((

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  • Address book with C programming; cannot compile my code.

    - by osabri
    I've divided my code into small programs so it can be easy to excute /* ab_error.c : in case of errors following messages will be displayed */ #include "adressbook.h" static char *errormsg[] = { "", "\nNot enough space on disk", "\nCannot open file", "\nCannot read file", "\nCannot write file" }; void check(int error) { switch(error) { case 0: return; case 1: write_file(); case 2: case 3: case 4: system("cls"); fputs(errormsg[error], stderr); exit(error); } } 2nd /* ab_fileio.c : functions for file input/output */ #include "adressbook.h" static char ab_file[] = "ADRESSBOOK.DAT"; //file to save the entries int read_file(void) { int error = 0; FILE *fp; ELEMENT *new_e, *last_e = NULL; DATA buffer; if( (fp = fopen(ab_file, "rb")) == NULL) return -1; //no file found while (fread(&buffer, sizeof(DATA), 1, fp) == 1) //reads one list element after another { if( (new_e = make_element()) == NULL) { error = 1; break; //not enough space } new_e->person = buffer; //copy data to new element new_e->next = NULL; if(hol.first == NULL) //list is empty? hol.first = new_e; //yes else last_e->next = new_e; //no last_e = new_e; ++hol.amount; } if( !error && !feof(fp) ) error = 3; //cannot read file fclose(fp); return error; } /*-------------------------------*/ int write_file(void) { int error = 0; FILE *fp; ELEMENT *p; if( (p = hol.first) == NULL) return 0; //list is empty if( (fp = fopen(ab_file, "wb")) == NULL) return 2; //cannot open while( p!= NULL) { if( fwrite(&p->person, sizeof(DATA), 1, fp) < 1) { error = 4; break; //cannot write } p = p->next; } fclose(fp); return error; } 3rd /* ab_list.c : functions to manipulate the list */ #include "adressbook.h" HOL hol = {0, NULL}; //global definition for head of list /* -------------------- */ ELEMENT *make_element(void) { return (ELEMENT *)malloc( sizeof(ELEMENT) ); } /* -------------------- */ int ins_element( DATA *newdata) { ELEMENT *new_e, *pre_p; if((new_e = make_element()) == NULL) return 1; new_e ->person = *newdata; // copy data to new element pre_p = search(new_e->person.family_name); if(pre_p == NULL) //no person in list { new_e->next = hol.first; //put it to the begin hol.first = new_e; } else { new_e->next = pre_p->next; pre_p->next = new_e; } ++hol.amount; return 0; } int erase_element( char name, char surname ) { return 0; } /* ---------------------*/ ELEMENT *search(char *name) { ELEMENT *sp, *retp; //searchpointer, returnpointer retp = NULL; sp = hol.first; while(sp != NULL && sp->person.family_name != name) { retp = sp; sp = sp->next; } return(retp); } 4th /* ab_screen.c : functions for printing information on screen */ #include "adressbook.h" #include <conio.h> #include <ctype.h> /* standard prompts for in- and output */ static char pgmname[] = "---- Oussama's Adressbook made in splendid C ----"; static char options[] = "\ 1: Enter new adress\n\n\ 2: Delete entry\n\n\ 3: Change entry\n\n\ 4: Print adress\n\n\ Esc: Exit\n\n\n\ Your choice . . .: "; static char prompt[] = "\ Name . . . .:\n\ Surname . . :\n\n\ Street . . .:\n\n\ House number:\n\n\ Postal code :\n\n\ Phone number:"; static char buttons[] = "\ <Esc> = cancel input <Backspace> = correct input\ <Return> = assume"; static char headline[] = "\ Name Surname Street House Postal code Phone number \n\ ------------------------------------------------------------------------"; static char further[] = "\ -------- continue with any key --------"; /* ---------------------------------- */ int menu(void) //show menu and read user input { int c; system ("cls"); set_cur(0,20); puts(pgmname); set_cur(6,0); printf("%s", options); while( (c = getch()) != ESC && (c < '1' || c > '4')) putch('\a'); return c; } /* ---------------------------------- */ int print_adr_book(void) //display adressbook { int line = 1; ELEMENT *p = hol.first; system("cls"); set_cur(0,20); puts(pgmname); set_cur(2,0); puts(headline); set_cur(5,0); while(p != NULL) //run through list and show entries { printf("%5d %-15s ",line, p->person.family_name); printf("%-12s %-15s ", p->person.given_name, p->person.street); printf("%-4d %-5d %-12d\n",p->person.house_number, p->person.postal_code, p->person.phone); p = p->next; if( p == NULL || ++line %16 == 1) //end of list or screen is full { set_cur(24,0); printf("%s",further); if( getch() == ESC) return 0; set_cur(5,0); scroll_up(0,5,24);//puts(headline); } } return 0; } /* -------------------------------------------*/ int make_entry(void) { char cache[50]; DATA newperson; ELEMENT *p; while(1) { system("cls"); set_cur(0,20); puts(pgmname); set_cur(6,0); puts("Please enter new data:"); set_cur(10,0); puts(prompt); set_cur(24,0); printf("%s",buttons); balken(10, 25, MAXL, ' ',0x70); //input name if(input(newperson.family_name, MAXL, ESC, CR) == ESC) return 0; balken(12,25, MAXL, ' ', 0x70); //surname if(input(newperson.given_name, MAXL, ESC, CR) == ESC) return 0; balken(14,25, 30, ' ', 0x70); //street if(input(newperson.street, 30, ESC, CR) == ESC) return 0; balken(16,25, 4, ' ',0x70); //housenumber if(input(cache, 4, ESC, CR) == ESC) return 0; newperson.house_number = atol(cache); //to string balken(18,25, 5, ' ',0x70); //postal code if(input(cache, 5, ESC, CR) == ESC) return 0; newperson.postal_code = atol(cache); //to string balken(20,25, 20, ' ',0x70); //phone number if(input(cache, 20, ESC, CR) == ESC) return 0; newperson.phone = atol(cache); //to string p = search(newperson.phone); if( p!= NULL && p->person.phone == newperson.phone) { set_cur(22,25); puts("phonenumber already exists!"); set_cur(24,0); printf("%s, further"); getch(); continue; } } } 5th /* adress_book_project.c : main program to create an adressbook */ /* copyrights by Oussama Sabri, June 2010 */ #include "adressbook.h" //project header file int main() { int rv, cmd; //return value, user command if ( (rv = read_file() ) == -1) // no data saved yet rv = make_entry(); check(rv); //prompts an error and quits program on disfunction do { switch (cmd = menu())//calls menu and gets user input back { case '1': rv = make_entry(); break; case '2': //delete entry case '3': //changes entry rv = change_entry(cmd); break; case '4': //prints adressbook on screen rv = print_adr_book(); break; case ESC: //end of program system ("cls"); rv = 0; break; } }while(cmd!= ESC); check ( write_file() ); //save adressbook return 0; } 6th /* Getcb.c --> Die Funktion getcb() liefert die naechste * * Tastatureingabe (ruft den BIOS-INT 0x16 auf). * * Return-Wert: * * ASCII-Code bzw. erweiterter Code + 256 */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> int getcb(void) { union REGS intregs; intregs.h.ah = 0; // Subfunktion 0: ein Zeichen // von der Tastatur lesen. int86( 0x16, &intregs, &intregs); if( intregs.h.al != 0) // Falls ASCII-Zeichen, return (intregs.h.al); // dieses zurueckgeben. else // Sonst den erweiterten return (intregs.h.ah + 0x100); // Code + 256 } 7th /* PUTCB.C --> enthaelt die Funktionen * * - putcb() * * - putcb9() * * - balken() * * - input() * * * * Es werden die Funktionen 9 und 14 des Video-Interrupts * * (ROM-BIOS-Interrupt 0x10) verwendet. * * * * Die Prototypen dieser Funktionen stehen in BIO.H */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> #define VIDEO_INT 0x10 /*---------------------------------------------------------------- * putcb(c) gibt das Zeichen auf der aktuellen Cursor-Position * am Bildschirm aus. Der Cursor wird versetzt. * Steuerzeichen Back-Space, CR, LF und BELL werden * ausgefuehrt. * Return-Wert: keiner */ void putcb(unsigned char c) /* Gibt das Zeichen in c auf */ { /* den Bildschirm aus. */ union REGS intregs; intregs.h.ah = 14; /* Subfunktion 14 ("Teletype") */ intregs.h.al = c; intregs.h.bl = 0xf; /* Vordergrund-Farbe im */ /* Grafik-Modus. */ int86(VIDEO_INT, &intregs, &intregs); } /*---------------------------------------------------------------- * putcb9(c,count,mode) gibt das Zeichen in c count-mal im * angegebenen Modus auf der aktuellen * Cursor-Position am Bildschirm aus. * Der Cursor wird nicht versetzt. * * Return-Wert: keiner */ void putcb9( unsigned char c, /* das Zeichen */ unsigned count, /* die Anzahl */ unsigned mode ) /* Low-Byte: das Atrribut */ { /* High-Byte: die Bildschirmseite*/ union REGS intregs; intregs.h.ah = 9; /* Subfunktion 9 des Int 0x10 */ intregs.h.al = c; intregs.x.bx = mode; intregs.x.cx = count; int86( VIDEO_INT, &intregs, &intregs); } /*---------------------------------------------------------------- * balken() positioniert den Cursor und zeichnet einen Balken, * wobei Position, L„nge, Fllzeichen und Attribut * als Argumente bergeben werden. * Der Cursor bleibt auf der ersten Position im Balken. */ void balken( unsigned int zeile, /* Start-Position */ unsigned int spalte, unsigned int laenge, /* Laenge des Balkens */ unsigned char c, /* Fuellzeichen */ unsigned int modus) /* Low-Byte: Attribut */ /* High-Byte: Bildschirmseite */ { union REGS intregs; intregs.h.ah = 2; /* Cursor auf der angegebenen */ intregs.h.dh = zeile; /* Bildschirmseite versetzen. */ intregs.h.dl = spalte; intregs.h.bh = (modus >> 8); int86(VIDEO_INT, &intregs, &intregs); putcb9(c, laenge, modus); /* Balken ausgeben. */ } /*---------------------------------------------------------------- * input() liest Zeichen von der Tastatur ein und haengt '\0' an. * Mit Backspace kann die Eingabe geloescht werden. * Das Attribut am Bildschirm bleibt erhalten. * * Argumente: 1. Zeiger auf den Eingabepuffer. * 2. Anzahl maximal einzulesender Zeichen. * 3. Die optionalen Argumente: Zeichen, mit denen die * Eingabe abgebrochen werden kann. * Diese Liste muá mit CR = '\r' enden! * Return-Wert: Das Zeichen, mit dem die Eingabe abgebrochen wurde. */ #include <stdarg.h> int getcb( void); /* Zum Lesen der Tastatur */ int input(char *puffer, int max,... ) { int c; /* aktuelles Zeichen */ int breakc; /* Abruchzeichen */ int nc = 0; /* Anzahl eingelesener Zeichen */ va_list argp; /* Zeiger auf die weiteren Arumente */ while(1) { *puffer = '\0'; va_start(argp, max); /* argp initialisieren */ c = getcb(); do /* Mit Zeichen der Abbruchliste vergleichen */ if(c == (breakc = va_arg(argp,int)) ) return(breakc); while( breakc != '\r' ); va_end( argp); if( c == '\b' && nc > 0) /* Backspace? */ { --nc; --puffer; putcb(c); putcb(' '); putcb(c); } else if( c >= 32 && c <= 255 && nc < max ) { ++nc; *puffer++ = c; putcb(c); } else if( nc == max) putcb('\7'); /* Ton ausgeben */ } } 8th /* Video.c --> Enthaelt die Funktionen * cls(), * scroll_up(), scroll_down(), * set_cur(), get_cur(), * set_screen_page(), get_screen_page() * * Die Prototypen dieser Funktionen befinden sich in BIO.H */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> #include "bio.h" #define VIDEO_INT 0x10 typedef unsigned char BYTE; void scroll_up( int anzahl, int anf_zeile, int end_zeile) { /* Fenster hoch rollen. */ union REGS intregs; intregs.x.ax = 0x600 + anzahl; /* Subfunktion AH = 6, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void scroll_down( int anzahl, int anf_zeile, int end_zeile) { /* Fenster runter rollen. */ union REGS intregs; intregs.x.ax = 0x700 + anzahl; /* Subfunktion AH = 7, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void set_cur( int zeile, int spalte) /* versetzt den Cursor */ { /* der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 2; intregs.h.dh = (BYTE)zeile; intregs.h.dl = (BYTE)spalte; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); } void get_cur(int *zeile, int *spalte) /* holt die Cursor- */ { /* Position der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 3; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); *zeile = (unsigned)intregs.h.dh; *spalte = (unsigned)intregs.h.dl; } void cls(void) { scroll_up(0,0,24); /* Gesamten Bildschirm loeschen. */ set_cur(0,0); /* Cursor in Home-Position. */ } int get_screen_page(void) /* Aktuelle Bildschirmseite holen.*/ { union REGS intregs; intregs.h.ah = 15; /* Subfunktion AH = 15: */ /* Bildschirm-Modus feststellen. */ int86(VIDEO_INT, &intregs, &intregs); return (intregs.h.bh); } void set_screen_page(int seite) /* setzt die aktive Seite des */ { /* Bildschirmpuffers auf die */ /* angegebene Seite. */ union REGS intregs; intregs.x.ax = 0x500 + seite; /* Subfunktion AH = 5 */ int86(VIDEO_INT, &intregs, &intregs); } /* ------------------------------------------------------------- Ein kleines Testprogramm : */ /* #include <stdio.h> int main() { cls(); set_cur(23, 0); printf("Weiter mit <Return>\n"); set_cur(12, 20); printf("Ein Test!\n"); getchar(); scroll_up(3, 5, 20); getchar(); scroll_down(6, 5, 20); getchar(); set_screen_page(1); printf("\nAuf der 2. Seite !\n"); getchar(); set_screen_page(0); set_cur(0,0); printf("\nWieder auf der 1. Seite !\n"); getchar(); cls(); return 0; } */ /* Video.c --> Enthaelt die Funktionen * cls(), * scroll_up(), scroll_down(), * set_cur(), get_cur(), * set_screen_page(), get_screen_page() * * Die Prototypen dieser Funktionen befinden sich in BIO.H */ /* Hinweis: Es muss ein DOS-Compiler verwendet werden. * * (z.B. der GNU-Compiler fuer DOS auf der CD) */ #include <dos.h> #include "bio.h" #define VIDEO_INT 0x10 typedef unsigned char BYTE; void scroll_up( int anzahl, int anf_zeile, int end_zeile) { /* Fenster hoch rollen. */ union REGS intregs; intregs.x.ax = 0x600 + anzahl; /* Subfunktion AH = 6, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void scroll_down( int anzahl, int anf_zeile, int end_zeile) { /* Fenster runter rollen. */ union REGS intregs; intregs.x.ax = 0x700 + anzahl; /* Subfunktion AH = 7, */ /* AL = Anzahl Zeilen. */ intregs.x.cx = anf_zeile << 8; /* CH=anf_zeile, cl=0 */ intregs.x.dx = (end_zeile <<8) | 79; /* DH=end_zeile,DL=79 */ intregs.h.bh = 7; /* normales Attribut */ int86(VIDEO_INT, &intregs, &intregs); } void set_cur( int zeile, int spalte) /* versetzt den Cursor */ { /* der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 2; intregs.h.dh = (BYTE)zeile; intregs.h.dl = (BYTE)spalte; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); } void get_cur(int *zeile, int *spalte) /* holt die Cursor- */ { /* Position der aktuellen Bildschirmseite.*/ union REGS intregs; intregs.h.ah = 3; intregs.h.bh = (BYTE)get_screen_page(); int86(VIDEO_INT, &intregs, &intregs); *zeile = (unsigned)intregs.h.dh; *spalte = (unsigned)intregs.h.dl; } void cls(void) { scroll_up(0,0,24); /* Gesamten Bildschirm loeschen. */ set_cur(0,0); /* Cursor in Home-Position. */ } int get_screen_page(void) /* Aktuelle Bildschirmseite holen.*/ { union REGS intregs; intregs.h.ah = 15; /* Subfunktion AH = 15: */ /* Bildschirm-Modus feststellen. */ int86(VIDEO_INT, &intregs, &intregs); return (intregs.h.bh); } void set_screen_page(int seite) /* setzt die aktive Seite des */ { /* Bildschirmpuffers auf die */ /* angegebene Seite. */ union REGS intregs; intregs.x.ax = 0x500 + seite; /* Subfunktion AH = 5 */ int86(VIDEO_INT, &intregs, &intregs); } /* ------------------------------------------------------------- Ein kleines Testprogramm : */ /* #include <stdio.h> int main() { cls(); set_cur(23, 0); printf("Weiter mit <Return>\n"); set_cur(12, 20); printf("Ein Test!\n"); getchar(); scroll_up(3, 5, 20); getchar(); scroll_down(6, 5, 20); getchar(); set_screen_page(1); printf("\nAuf der 2. Seite !\n"); getchar(); set_screen_page(0); set_cur(0,0); printf("\nWieder auf der 1. Seite !\n"); getchar(); cls(); return 0; } */ /* BIO.H --> Enthaelt die Prototypen der BIOS-Funktionen. */ /* --- Funktionen in VIDEO.C --- */ extern void scroll_up(int anzahl, int anf_zeile,int end_zeile); extern void scroll_down(int anzahl, int anf_zeile, int end_zeile); extern void set_cur(int zeile, int spalte); extern void get_cur(int *zeile, int *spalte); extern void cls(void); extern int get_screen_page(void); extern void set_screen_page(int page); /* --- Funktionen in GETCB.C / PUTCB.C --- */ extern int getcb(void); extern void putcb(int c); extern void putcb9(int c, unsigned count, unsigned modus); extern void balken(int zeile, int spalte, int laenge, int c, unsigned modus); extern int input(char *puffer, int max,... ); need your help, can't find my mistakes:((

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  • Problem in linking an nasm code

    - by Stefano
    I'm using a computer with an Intel Core 2 CPU and 2GB of RAM. The SO is Ubuntu 9.04. When I try to compile this code: ;programma per la simulazione di un terminale su PC, ottenuto utilizzando l'8250 ;in condizione di loopback , cioè Tx=Rx section .code64 section .data TXDATA EQU 03F8H ;TRASMETTITORE RXDATA EQU 03F8H ;RICEVITORE BAUDLSB EQU 03F8H ;DIVISORE DI BAUD RATE IN LSB BAUDMSB EQU 03F9H ;DIVISORE DI BAUD RATE IN MSB INTENABLE EQU 03F9H ;REGISTRO DI ABILITAZIONE DELL'INTERRUZIONE INTIDENTIF EQU 03FAH ;REGISTRO DI IDENTIFICAZIONE DELL'INTERRUZIONE LINECTRL EQU 03FBH ;REGISTRO DI CONTROLLO DELLA LINEA MODEMCTRL EQU 03FCH ;REGISTRO DI CONTROLLO DEL MODEM LINESTATUS EQU 03FDH ;REGISTRO DI STATO DELLA LINEA MODEMSTATUS EQU 03FEH ;REGISTRO DI STATO DEL MODEM BAUDRATEDIV DW 0060H ;DIVISOR: LOW=60, HIGH=00 -BAUD =9600 COUNTERCHAR DB 0 ;CHARACTER COUNTER ;DW 256 DUP (?) section .text global _start _start: ;PROGRAMMAZIONE 8250 MOV DX,LINECTRL MOV AL,80H ;BIT 7=1 PER INDIRIZZARE IL BAUD RATE OUT DX,AL MOV DX,BAUDLSB MOV AX,BAUDRATEDIV ;DEFINISCO FATTORE DI DIVISIONE OUT DX,AL MOV DX,BAUDMSB MOV AL,AH OUT DX,AL ;MSB MOV DX,LINECTRL MOV AL,00000011B ;8 BIT DATO, 1 STOP, PARITA' NO OUT DX,AL MOV DX,MODEMCTRL MOV AL,00010011B ;BIT 4=0 PER NO LOOPBACK OUT DX,AL MOV DX,INTENABLE XOR AL,AL ;DISABILITO TUTTI GLI INTERRUPTS OUT DX,AL CICLO: MOV DX,LINESTATUS IN AL,DX ;LEGGO IL REGISTRO DI STATO DELLA LINEA TEST AL,00011110B ;VERIFICO GLI ERRORI (4 TIPI) JNE ERRORI TEST AL,01H ;VERIFICO Rx PRONTO JNE LEGGOCHAR TEST AL,20H ;VERIFICO Tx VUOTO JE CICLO ;SE SI ARRIVA A QUESTO PUNTO ALLORA L'8250 è PRONTO PER TRASMETTERE UN NUOVO CARATTERE MOV AH,1 INT 80H JE CICLO ;SE SI ARRIVA A QUESTO PUNTO SIGNIFICA CHE ESISTE UN CARATTERE DA TASTIERA MOV AH,0 INT 80H ;Al CONTIENE IL CARATTERE DELLA TASTIERA MOV DX,3F8H OUT DX,AL JMP CICLO LEGGOCHAR: MOV AL,[COUNTERCHAR] INC AL CMP AL,15 JE FINE MOV [COUNTERCHAR],AL MOV DX,TXDATA IN AL,DX ;AL CONTIENE IL CARATTERE RICEVUTO AND AL,7FH ;POICHè VI SONO 7 BIT DI DATO ;VISUALIZZAZIONE DEL CARATTERE MOV BX,0 MOV AH,14 INT 80H POP AX CMP AL,0DH ;CONTROLLO SE RETURN JNE CICLO ;CAMBIO RIGA DI VISUALIZZAZIONE MOV AL,0AH MOV BX,0 MOV AH,14 ;INT 10H INT 80H JMP CICLO ;GESTIONE ERRORI ERRORI: MOV DX,3F8H IN AL,DX MOV AL,'?' MOV BX,0 MOV AH,14 INT 80H JMP CICLO FINE: XOR AH,AH MOV AL,03 INT 80H When I compile this code "NASM -f bin UARTLOOP.asm", the compiler can create the UARTLOOP.o file without any error. When I try to link the .o file with "ld UARTLOOP.o" it tells: UARTLOOP.o: In function `_start': UARTLOOP.asm:(.text+0xd): relocation truncated to fit: R_X86_64_16 against `.data' Have u got some ideas to solve this problem? Thx =)

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  • Output character in assembly

    - by lolopolosko
    Please help me! How i can output character that moves around the perimeter of a rectangle (10*5 or 15*7) in console with TASM? .MODEL small .STACK 100h .CODE start: mov ah,03 int 10h mov cx,10 A: push cx mov ah,03 int 10h mov ah,02h inc dl int 10h mov al,42 int 29h pop cx LOOP A mov ah,4ch int 21h end start I do not know how to solve the problem...

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  • Calculating the number of occasions with a set period sumproduct function

    - by user158056
    =SUMPRODUCT((F16:F274=("A")) *(F17:F275<>("A"))) +SUMPRODUCT((F16:F274=("AH")) *(F17:F275<>("AH"))) +SUMPRODUCT((F16:F274=("AU")) *(F17:F275<>("AU"))) +SUMPRODUCT((F16:F274=("AHU"))*(F17:F275<>("AHU"))) I am using the above formula to add the number of occasions sickness occurs using the following as a key. It works fine until you get say an A and a AH in the same sickness period. Instead of reporting just one occasion off it reports two. Is there a way I can separate this? Absence A Absence 1/2 AH Absence Unpaid AU Absence 1/2 Unpaid AHU

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  • ASM programming, how to use loop?

    - by chris
    Hello. Im first time here.I am a college student. I've created a simple program by using assembly language. And im wondering if i can use loop method to run it almost samething as what it does below the program i posted. and im also eager to find someome who i can talk through MSN messanger so i can ask you questions right away.(if possible) ok thank you .MODEL small .STACK 400h .data prompt db 10,13,'Please enter a 3 digit number, example 100:',10,13,'$' ;10,13 cause to go to next line first_digit db 0d second_digit db 0d third_digit db 0d Not_prime db 10,13,'This number is not prime!',10,13,'$' prime db 10,13,'This number is prime!',10,13,'$' question db 10,13,'Do you want to contine Y/N $' counter dw 0d number dw 0d half dw ? .code Start: mov ax, @data ;establish access to the data segment mov ds, ax mov number, 0d LetsRoll: mov dx, offset prompt ; print the string (please enter a 3 digit...) mov ah, 9h int 21h ;execute ;read FIRST DIGIT mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov first_digit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ;AND that the result will use both dx::ax ;but we understand that dx will contain only leading zeros add number, ax ;save ;variable <number> now contains 1st digit * 10 ;---------------------------------------------------------------------- ;read SECOND DIGIT, multiply by 10 and add in mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov second_digit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;AND that the result will use both dx::ax ;but we understand that dx will contain only leading zeros. Ignore them add number, ax ;save -- nearly finished ;variable <number> now contains 1st digit * 100 + second digit * 10 ;---------------------------------------------------------------------- ;read THIRD DIGIT, add it in (no multiplication this time) mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov third_digit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer add number, ax ;Both my variable number and ax are 16 bits, so equal size mov ax, number ;copy contents of number to ax mov cx, 2h div cx ;Divide by cx mov half, ax ;copy the contents of ax to half mov cx, 2h; mov ax, number; ;copy numbers to ax xor dx, dx ;flush dx jmp prime_check ;jump to prime check print_question: mov dx, offset question ;print string (do you want to continue Y/N?) mov ah, 9h int 21h ;execute mov ah, 1h int 21h ;execute cmp al, 4eh ;compare je Exit ;jump to exit cmp al, 6eh ;compare je Exit ;jump to exit cmp al, 59h ;compare je Start ;jump to start cmp al, 79h ;compare je Start ;jump to start prime_check: div cx; ;Divide by cx cmp dx, 0h ;reset the value of dx je print_not_prime ;jump to not prime xor dx, dx; ;flush dx mov ax, number ;copy the contents of number to ax cmp cx, half ;compare half with cx je print_prime ;jump to print prime section inc cx; ;increment cx by one jmp prime_check ;repeat the prime check print_prime: mov dx, offset prime ;print string (this number is prime!) mov ah, 9h int 21h ;execute jmp print_question ;jumps to question (do you want to continue Y/N?) this is for repeat print_not_prime: mov dx, offset Not_prime ;print string (this number is not prime!) mov ah, 9h int 21h ;execute jmp print_question ;jumps to question (do you want to continue Y/N?) this is for repeat Exit: mov ah, 4ch int 21h ;execute exit END Start

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  • CUDA: cudaMemcpy only works in emulation mode.

    - by Jason
    I am just starting to learn how to use CUDA. I am trying to run some simple example code: float *ah, *bh, *ad, *bd; ah = (float *)malloc(sizeof(float)*4); bh = (float *)malloc(sizeof(float)*4); cudaMalloc((void **) &ad, sizeof(float)*4); cudaMalloc((void **) &bd, sizeof(float)*4); ... initialize ah ... /* copy array on device */ cudaMemcpy(ad,ah,sizeof(float)*N,cudaMemcpyHostToDevice); cudaMemcpy(bd,ad,sizeof(float)*N,cudaMemcpyDeviceToDevice); cudaMemcpy(bh,bd,sizeof(float)*N,cudaMemcpyDeviceToHost); When I run in emulation mode (nvcc -deviceemu) it runs fine (and actually copies the array). But when I run it in regular mode, it runs w/o error, but never copies the data. It's as if the cudaMemcpy lines are just ignored. What am I doing wrong? Thank you very much, Jason

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  • reading from File in assembly

    - by Natasha
    i am trying to read a username and a password from a file in x86 assembly for the perpose of authentication obviously the file consists of two lines , the user name and the password how can i read the two lines seperately and compare them? My attempt: proc read_file mov ah,3dh lea dx,file_name int 21h mov bx, ax xor si,si repeat: mov ah, 3fh lea dx, buffer mov cx, 100 int 21h mov si, ax mov buffer[si], '$' mov ah, 09h int 21h ;print on screen cmp si, 100 je repeat jmp stop;jump to end stop: RET read_file ENDP

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  • Illegal instruction in Assembly

    - by Natasha
    I really do not understand why this simple code works fine in the first attempt but when putting it in a procedure an error shows: NTVDM CPU has encountered an illegal instruction CS:db22 IP:4de4 OP:f0 ff ff ff ff The first code segment works just fine: .model small .stack 100h .code start: mov ax,@data mov ds,ax mov es,ax MOV AH,02H ;sets cursor up MOV BH,00H MOV DH,02 MOV DL,00 INT 10H EXIT: MOV AH,4CH INT 21H END However This generates an error: .model small .stack 100h .code start: mov ax,@data mov ds,ax mov es,ax call set_cursor PROC set_cursor near MOV AH,02H ;sets cursor up MOV BH,00H MOV DH,02 MOV DL,00 INT 10H RET set_cursor ENDP EXIT: MOV AH,4CH INT 21H END Note: Nothing is wrong with windows config. I have tried many sample codes that work fine Thanks

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  • Problem with asm program (nasm)

    - by GLeBaTi
    org 0x100 SEGMENT .CODE mov ah,0x9 mov dx, Msg1 int 0x21 ;string input mov ah,0xA mov dx,buff int 0x21 mov ax,0 mov al,[buff+1]; length ;string UPPERCASE mov cl, al mov si, buff cld loop1: lodsb; cmp al, 'a' jnb upper loop loop1 ;output mov ah,0x9 mov dx, buff int 0x21 exit: mov ah, 0x8 int 0x21 int 0x20 upper: sub al,32 jmp loop1 SEGMENT .DATA Msg1 db 'Press string: $' buff db 254,0 this code perform poorly. I think that problem in "jnb upper". This program make small symbols into big symbols.

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  • easy asm program(nasm)

    - by GLeBaTi
    org 0x100 SEGMENT .CODE mov ah,0x9 mov dx, Msg1 int 0x21 ;string input mov ah,0xA mov dx,buff int 0x21 mov ax,0 mov al,[buff+1]; length ;string UPPERCASE mov cl, al mov si, buff cld loop1: lodsb; cmp al, 'a' jnb upper loop loop1 ;output mov ah,0x9 mov dx, buff int 0x21 exit: mov ah, 0x8 int 0x21 int 0x20 upper: sub al,32 jmp loop1 SEGMENT .DATA Msg1 db 'Press string: $' buff db 254,0 this code perform poorly. I think that problem in "jnb upper". This program make small symbols into big symbols.

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  • Scan for first zero bit (Assembly)?

    - by cthulhu
    I have some numbers in AH, AL, BL, BH registers. I need to check whether there is 0 bit in each of the registers in left half of the number. If yes, then put into check variable value 10 else -10. How can I do this? I tried something like that: org 100h check dw 0 mov ah, 11111111b mov al, 11111111b mov bl, 11111111b mov bh, 11111111b mov check, -10 shr ah, 4 shr al, 4 shr bl, 4 shr bh, 4 cmp ah, 0Fh jz first first: cmp al, 0Fh jz second second: cmp bl, 0Fh jz third third: cmp bh, 0Fh jz final final: mov check, 10 ret

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  • Getting to grips with the stack in nasm

    - by MarkPearl
    Today I spent a good part of my day getting to grips with the stack and nasm. After looking at my notes on nasm I think this is one area for the course I am doing they could focus more on… So here are some snippets I have put together that have helped me understand a little bit about the stack… Simplest example of the stack You will probably see examples like the following in circulation… these demonstrate the simplest use of the stack… org 0x100 bits 16 jmp main main: push 42h push 43h push 44h mov ah,2h ;set to display characters pop dx    ;get the first value int 21h   ;and display it pop dx    ;get 2nd value int 21h   ;and display it pop dx    ;get 3rd value int 21h   ;and display it int 20h The output from above code would be… DCB Decoupling code using “call” and “ret” This is great, but it oversimplifies what I want to use the stack for… I do not know if this goes against the grain of assembly programmers or not, but I want to write loosely coupled assembly code – and I want to use the stack as a mechanism for passing values into my decoupled code. In nasm we have the call and return instructions, which provides a mechanism for decoupling code, for example the following could be done… org 0x100 bits 16 jmp main ;---------------------------------------- displayChar: mov ah,2h mov dx,41h int 21h ret ;---------------------------------------- main: call displayChar int 20h   This would output the following to the console A So, it would seem that call and ret allow us to jump to segments of our code and then return back to the calling position – a form of segmenting the code into what we would called in higher order languages “functions” or “methods”. The only issue is, in higher order languages there is a way to pass parameters into the functions and return results. Because of the primitive nature of the call and ret instructions, this does not seem to be obvious. We could of course use the registers to pass values into the subroutine and set values coming out, but the problem with this is we… Have a limited number of registers Are threading our code with tight coupling (it would be hard to migrate methods outside of their intended use in a particular program to another one) With that in mind, I turn to the stack to provide a loosely coupled way of calling subroutines… First attempt with the Stack Initially I thought this would be simple… we could use code that looks as follows to achieve what I want… org 0x100 bits 16 jmp main ;---------------------------------------- displayChar: mov ah,2h pop dx int 21h ret ;---------------------------------------- main: push 41h call displayChar int 20h   However running this application does not give the desired result, I want an ‘A’ to be returned, and I am getting something totally different (you will to). Reading up on the call and ret instructions a discovery is made… they are pushing and popping things onto and off the stack as well… When the call instruction is executed, the current value of IP (the address of the instruction to follow) is pushed onto the stack, when ret is called, the last value on the stack is popped off into the IP register. In effect what the above code is doing is as follows with the stack… push 41h push current value of ip pop current value of ip to dx pop 41h to ip This is not what I want, I need to access the 41h that I pushed onto the stack, but the call value (which is necessary) is putting something in my way. So, what to do? Remember we have other registers we can use as well as a thing called indirect addressing… So, after some reading around, I came up with the following approach using indirect addressing… org 0x100 bits 16 jmp main ;---------------------------------------- displayChar: mov bp,sp mov ah,2h mov dx,[bp+2] int 21h ret ;---------------------------------------- main: push 41h call displayChar int 20h In essence, what I have done here is used a trick with the stack pointer… it goes as follows… Push 41 onto the stack Make the call to the function, which will push the IP register onto the stack and then jump to the displayChar label Move the value in the stack point to the bp register (sp currently points at IP register) Move the at the location of bp minus 2 bytes to dx (this is now the value 41h) display it, execute the ret instruction, which pops the ip value off the stack and goes back to the calling point This approach is still very raw, some further reading around shows that I should be pushing the value of bp onto the stack before replacing it with sp, but it is the starting thread to getting loosely coupled subroutines. Let’s see if you get what the following output would be? org 0x100 bits 16 jmp main ;---------------------------------------- displayChar: mov bp,sp mov ah,2h mov dx,[bp+4] int 21h mov dx,[bp+2] int 21h ret ;---------------------------------------- main: push 41h push 42h call displayChar int 20h The output is… AB Where to from here? If by any luck some assembly programmer comes along and see this code and notices that I have made some fundamental flaw in my logic… I would like to know, so please leave a comment… appreciate any feedback!

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  • Assembly load and execute issue

    - by Jean Carlos Suárez Marranzini
    I'm trying to develop Assembly code allowing me to load and execute(by input of the user) 2 other Assembly .EXE programs. I'm having two problems: -I don't seem to be able to assign the pathname to a valid register(Or maybe incorrect syntax) -I need to be able to execute the other program after the first one (could be either) started its execution. This is what I have so far: mov ax,cs ; moving code segment to data segment mov ds,ax mov ah,1h ; here I read from keyboard int 21h mov dl,al cmp al,'1' ; if 1 jump to LOADRUN1 JE LOADRUN1 popf cmp al,'2' ; if 1 jump to LOADRUN2 JE LOADRUN2 popf LOADRUN1: MOV AH,4BH MOV AL,00 LEA DX,[PROGNAME1] ; Not sure if it works INT 21H LOADRUN2: MOV AH,4BH MOV AL,00 LEA DX,[PROGNAME2] ; Not sure if it works INT 21H ; Here I define the bytes containing the pathnames PROGNAME1 db 'C:\Users\Usuario\NASM\Adding.exe',0 PROGNAME2 db 'C:\Users\Usuario\NASM\Substracting.exe',0 I just don't know how start another program by input in the 'parent' program, after one is already executing. Thanks in advance for your help! Any additional information I'll be more than happy to provide. -I'm using NASM 16 bits, Windows 7 32 bits.

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  • image upload problem.

    - by Bluemagica
    I wrote a function to resize and upload images...it works, but only for one image. So if I call the function three times, I end up with 3 copies of the last image..... function uploadImage($name,$width,$height,$size,$path='content/user_avatars/') { //=================================================== //Handle image upload $upload_error=0; //Picture $img = $_FILES[$name]['name']; if($img) { $file = stripslashes($_FILES[$name]['name']); $ext = strtolower(getExt($file)); if($ext!='jpg' && $ext!='jpeg' && $ext!='png' && $ext!='gif') { $error_msg = "Unknown extension"; $upload_error = 1; return array($upload_error,$error_msg); } if(filesize($_FILES[$name]['tmp_name'])>$size*1024) { $error_msg = "Max file size of ".($size*1024)."kb exceeded"; $upload_error = 2; return array($upload_error,$error_msg); } $newFile = time().'.'.$ext; resizeImg($_FILES[$name]['tmp_name'],$ext,$width,$height); $store = copy($_FILES[$name]['tmp_name'],$path.$newFile); if(!$store) { $error_msg = "Uploading failed"; $upload_error = 3; return array($upload_error,$error_msg); } else { return array($upload_error,$newFile); } } } //========================================================================================= //Helper Functions function getExt($str) { $i = strpos($str,"."); if(!$i) { return ""; } $l = strlen($str)-$i; $ext = substr($str,$i+1,$l); return $ext; } function resizeImg($file,$ext,$width,$height) { list($aw,$ah) = getimagesize($file); $scaleX = $aw/$width; $scaleY = $ah/$height; if($scaleX>$scaleY) { $nw = round($aw*(1/$scaleX)); $nh = round($ah*(1/$scaleX)); } else { $nw = round($aw*(1/$scaleY)); $nh = round($ah*(1/$scaleY)); } $new_image = imagecreatetruecolor($nw,$nh); imagefill($new_image,0,0,imagecolorallocatealpha($new_image,255,255,255,127)); if($ext=='jpg'||$ext=='jpeg') { $src_image = imagecreatefromjpeg($file); } else if($ext=='gif') { $src_image = imagecreatefromgif($file); } else if($ext=='png') { $src_image = imagecreatefrompng($file); } imagecopyresampled($new_image,$src_image,0,0,0,0,$nw,$nh,$aw,$ah); if($ext=='jpg'||$ext=='jpeg') { imagejpeg($new_image,$file,100); } else if($ext=='gif') { imagegif($new_image,$file); } else if($ext=='png') { imagepng($new_image,$file,9); } imagedestroy($src_image); imagedestroy($new_image); } I have a form with two upload fields, 'face_pic' and 'body_pic', and I want to upload these two to the server and resize them before storing. Any ideas?

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