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  • Syntactical analysis with Flex/Bison part 2

    - by Imran
    Hallo, I need help in Lex/Yacc Programming. I wrote a compiler for a syntactical analysis for inputs of many statements. Now i have a special problem. In case of an Input the compiler gives the right output, which statement is uses, constant operator or a jmp instructor to which label, now i have to write so, if now a if statement comes, first the first command (before the else) must be give out when the assignment of the if is yes then it must jump to the end because the command after the else isnt needed, so after this jmp then the second command must be give out. I show it in an example maybe you understand what i mean. Input adr. Output if(x==0) 10 if(x==0) Wait 5 20 WAIT 5 else 30 JMP 50 Wait 1 40 WAIT 1 end 50 END like so. I have an idea, maybe i can do it whith a special if statement like IF exp jmp_stmt_end stmt_seq END when the if statement is given in the input the compiler has to recognize the end ofthe statement and like my jmp_stmt in my compiler ( you have to download the files from http://bitbucket.org/matrix/changed-tiny) only to jump to the end. I hope you understand my problem.thanks.

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  • nasm infinite loop with FPU

    - by Ben Ishak
    i'm trying to create a small nasm program which do this operation in floating point while(input <= 10^5) do begin input = input * 10 i = i - 1 end the equivilant program in nasm is as following section .data input: resd 1 n10: dd 0x41200000 ; 10 _start: mov eax, 200 ; eax = 200 ; extract eax -> Floating Point IEEE 754 and eax, 0x7f800000 shr eax, 23 sub eax, 127 mov dword [input], eax ; input = eax = 200 mov edx, 0x49742400 ; 10^5 ; %begin mov ecx, 0 ; i = 0 jmp alpha alpha: fld dword [input] cmp [input], edx ; input <= 10^5 jle _while jmp log2 _while: fld dword [n10] ; 10 fld dword [input] ; input fmul st0, st1 ; input * 10 fst dword [input] ; input = input dec ecx ; i = i - 1 jmp alpha the _while loop is iterating infinitely ecx / i gards always the same value = -4194304 (it is sepposed to be 0) and doesn't decrement

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  • The Java Specialist: An Interview with Java Champion Heinz Kabutz

    - by Janice J. Heiss
    Dr. Heinz Kabutz is well known for his Java Specialists’ Newsletter, initiated in November 2000, where he displays his acute grasp of the intricacies of the Java platform for an estimated 70,000 readers; for his work as a consultant; and for his workshops and trainings at his home on the Island of Crete where he has lived since 2006 -- where he is known to curl up on the beach with his laptop to hack away, in between dips in the Mediterranean. Kabutz was born of German parents and raised in Cape Town, South Africa, where he developed a love of programming in junior high school through his explorations on a ZX Spectrum computer. He received a B.S. from the University of Cape Town, and at 25, a Ph.D., both in computer science. He will be leading a two-hour hands-on lab session, HOL6500 – “Finding and Solving Java Deadlocks,” at this year’s JavaOne that will explore what causes deadlocks and how to solve them. Q: Tell us about your JavaOne plans.A: I am arriving on Sunday evening and have just one hands-on-lab to do on Monday morning. This is the first time that a non-Oracle team is doing a HOL at JavaOne under Oracle's stewardship and we are all a bit nervous about how it will turn out. Oracle has been immensely helpful in getting us set up. I have a great team helping me: Kirk Pepperdine, Dario Laverde, Benjamin Evans and Martijn Verburg from jClarity, Nathan Reynolds from Oracle, Henri Tremblay of OCTO Technology and Jeff Genender of Savoir Technologies. Monday will be hard work, but after that, I will hopefully get to network with fellow Java experts, attend interesting sessions and just enjoy San Francisco. Oh, and my kids have already given me a shopping list of things to get, like a GoPro Hero 2 dive housing for shooting those nice videos of Crete. (That's me at the beginning diving down.) Q: What sessions are you attending that we should know about?A: Sometimes the most unusual sessions are the best. I avoid the "big names". They often are spread too thin with all their sessions, which makes it difficult for them to deliver what I would consider deep content. I also avoid entertainers who might be good at presenting but who do not say that much.In 2010, I attended a session by Vladimir Yaroslavskiy where he talked about sorting. Although he struggled to speak English, what he had to say was spectacular. There was hardly anybody in the room, having not heard of Vladimir before. To me that was the highlight of 2010. Funnily enough, he was supposed to speak with Joshua Bloch, but if you remember, Google cancelled. If Bloch has been there, the room would have been packed to capacity.Q: Give us an update on the Java Specialists’ Newsletter.A: The Java Specialists' Newsletter continues being read by an elite audience around the world. The apostrophe in the name is significant.  It is a newsletter for Java specialists. When I started it twelve years ago, I was trying to find non-obvious things in Java to write about. Things that would be interesting to an advanced audience.As an April Fool's joke, I told my readers in Issue 44 that subscribing would remain free, but that they would have to pay US$5 to US$7 depending on their geographical location. I received quite a few angry emails from that one. I would have not earned that much from unsubscriptions. Most readers stay for a very long time.After Oracle bought Sun, the Java community held its breath for about two years whilst Oracle was figuring out what to do with Java. For a while, we were quite concerned that there was not much progress shown by Oracle. My newsletter still continued, but it was quite difficult finding new things to write about. We have probably about 70,000 readers, which is quite a small number for a Java publication. However, our readers are the top in the Java industry. So I don't mind having "only" 70000 readers, as long as they are the top 0.7%.Java concurrency is a very important topic that programmers think they should know about, but often neglect to fully understand. I continued writing about that and made some interesting discoveries. For example, in Issue 165, I showed how we can get thread starvation with the ReadWriteLock. This was a bug in Java 5, which was corrected in Java 6, but perhaps a bit too much. Whereas we could get starvation of writers in Java 5, in Java 6 we could now get starvation of readers. All of these interesting findings make their way into my courseware to help companies avoid these pitfalls.Another interesting discovery was how polymorphism works in the Server HotSpot compiler in Issue 157 and Issue 158. HotSpot can inline methods from interfaces that have only one implementation class in the JVM. When a new subclass is instantiated and called for the first time, the JVM will undo the previous optimization and re-optimize differently.Here is a little memory puzzle for your readers: public class JavaMemoryPuzzle {  private final int dataSize =      (int) (Runtime.getRuntime().maxMemory() * 0.6);  public void f() {    {      byte[] data = new byte[dataSize];    }    byte[] data2 = new byte[dataSize];  }  public static void main(String[] args) {    JavaMemoryPuzzle jmp = new JavaMemoryPuzzle();    jmp.f();  }}When you run this you will always get an OutOfMemoryError, even though the local variable data is no longer visible outside of the code block.So here comes the puzzle, that I'd like you to ponder a bit. If you very politely ask the VM to release memory, then you don't get an OutOfMemoryError: public class JavaMemoryPuzzlePolite {  private final int dataSize =      (int) (Runtime.getRuntime().maxMemory() * 0.6);  public void f() {    {      byte[] data = new byte[dataSize];    }    for(int i=0; i<10; i++) {      System.out.println("Please be so kind and release memory");    }    byte[] data2 = new byte[dataSize];  }  public static void main(String[] args) {    JavaMemoryPuzzlePolite jmp = new JavaMemoryPuzzlePolite();    jmp.f();    System.out.println("No OutOfMemoryError");  }}Why does this work? When I published this in my newsletter, I received over 400 emails from excited readers around the world, most of whom sent me the wrong explanation. After the 300th wrong answer, my replies became unfortunately a bit curt. Have a look at Issue 174 for a detailed explanation, but before you do, put on your thinking caps and try to figure it out yourself. Q: What do you think Java developers should know that they currently do not know?A: They should definitely get to know more about concurrency. It is a tough subject that most programmers try to avoid. Unfortunately we do come in contact with it. And when we do, we need to know how to protect ourselves and how to solve tricky system errors.Knowing your IDE is also useful. Most IDEs have a ton of shortcuts, which can make you a lot more productive in moving code around. Another thing that is useful is being able to read GC logs. Kirk Pepperdine has a great talk at JavaOne that I can recommend if you want to learn more. It's this: CON5405 – “Are Your Garbage Collection Logs Speaking to You?” Q: What are you looking forward to in Java 8?A: I'm quite excited about lambdas, though I must confess that I have not studied them in detail yet. Maurice Naftalin's Lambda FAQ is quite a good start to document what you can do with them. I'm looking forward to finding all the interesting bugs that we will now get due to lambdas obscuring what is really going on underneath, just like we had with generics.I am quite impressed with what the team at Oracle did with OpenJDK's performance. A lot of the benchmarks now run faster.Hopefully Java 8 will come with JSR 310, the Date and Time API. It still boggles my mind that such an important API has been left out in the cold for so long.What I am not looking forward to is losing perm space. Even though some systems run out of perm space, at least the problem is contained and they usually manage to work around it. In most cases, this is due to a memory leak in that region of memory. Once they bundle perm space with the old generation, I predict that memory leaks in perm space will be harder to find. More contracts for us, but also more pain for our customers. Originally published on blogs.oracle.com/javaone.

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  • The Java Specialist: An Interview with Java Champion Heinz Kabutz

    - by Janice J. Heiss
    Dr. Heinz Kabutz is well known for his Java Specialists’ Newsletter, initiated in November 2000, where he displays his acute grasp of the intricacies of the Java platform for an estimated 70,000 readers; for his work as a consultant; and for his workshops and trainings at his home on the Island of Crete where he has lived since 2006 -- where he is known to curl up on the beach with his laptop to hack away, in between dips in the Mediterranean. Kabutz was born of German parents and raised in Cape Town, South Africa, where he developed a love of programming in junior high school through his explorations on a ZX Spectrum computer. He received a B.S. from the University of Cape Town, and at 25, a Ph.D., both in computer science. He will be leading a two-hour hands-on lab session, HOL6500 – “Finding and Solving Java Deadlocks,” at this year’s JavaOne that will explore what causes deadlocks and how to solve them. Q: Tell us about your JavaOne plans.A: I am arriving on Sunday evening and have just one hands-on-lab to do on Monday morning. This is the first time that a non-Oracle team is doing a HOL at JavaOne under Oracle's stewardship and we are all a bit nervous about how it will turn out. Oracle has been immensely helpful in getting us set up. I have a great team helping me: Kirk Pepperdine, Dario Laverde, Benjamin Evans and Martijn Verburg from jClarity, Nathan Reynolds from Oracle, Henri Tremblay of OCTO Technology and Jeff Genender of Savoir Technologies. Monday will be hard work, but after that, I will hopefully get to network with fellow Java experts, attend interesting sessions and just enjoy San Francisco. Oh, and my kids have already given me a shopping list of things to get, like a GoPro Hero 2 dive housing for shooting those nice videos of Crete. (That's me at the beginning diving down.) Q: What sessions are you attending that we should know about?A: Sometimes the most unusual sessions are the best. I avoid the "big names". They often are spread too thin with all their sessions, which makes it difficult for them to deliver what I would consider deep content. I also avoid entertainers who might be good at presenting but who do not say that much.In 2010, I attended a session by Vladimir Yaroslavskiy where he talked about sorting. Although he struggled to speak English, what he had to say was spectacular. There was hardly anybody in the room, having not heard of Vladimir before. To me that was the highlight of 2010. Funnily enough, he was supposed to speak with Joshua Bloch, but if you remember, Google cancelled. If Bloch has been there, the room would have been packed to capacity.Q: Give us an update on the Java Specialists’ Newsletter.A: The Java Specialists' Newsletter continues being read by an elite audience around the world. The apostrophe in the name is significant.  It is a newsletter for Java specialists. When I started it twelve years ago, I was trying to find non-obvious things in Java to write about. Things that would be interesting to an advanced audience.As an April Fool's joke, I told my readers in Issue 44 that subscribing would remain free, but that they would have to pay US$5 to US$7 depending on their geographical location. I received quite a few angry emails from that one. I would have not earned that much from unsubscriptions. Most readers stay for a very long time.After Oracle bought Sun, the Java community held its breath for about two years whilst Oracle was figuring out what to do with Java. For a while, we were quite concerned that there was not much progress shown by Oracle. My newsletter still continued, but it was quite difficult finding new things to write about. We have probably about 70,000 readers, which is quite a small number for a Java publication. However, our readers are the top in the Java industry. So I don't mind having "only" 70000 readers, as long as they are the top 0.7%.Java concurrency is a very important topic that programmers think they should know about, but often neglect to fully understand. I continued writing about that and made some interesting discoveries. For example, in Issue 165, I showed how we can get thread starvation with the ReadWriteLock. This was a bug in Java 5, which was corrected in Java 6, but perhaps a bit too much. Whereas we could get starvation of writers in Java 5, in Java 6 we could now get starvation of readers. All of these interesting findings make their way into my courseware to help companies avoid these pitfalls.Another interesting discovery was how polymorphism works in the Server HotSpot compiler in Issue 157 and Issue 158. HotSpot can inline methods from interfaces that have only one implementation class in the JVM. When a new subclass is instantiated and called for the first time, the JVM will undo the previous optimization and re-optimize differently.Here is a little memory puzzle for your readers: public class JavaMemoryPuzzle {  private final int dataSize =      (int) (Runtime.getRuntime().maxMemory() * 0.6);  public void f() {    {      byte[] data = new byte[dataSize];    }    byte[] data2 = new byte[dataSize];  }  public static void main(String[] args) {    JavaMemoryPuzzle jmp = new JavaMemoryPuzzle();    jmp.f();  }}When you run this you will always get an OutOfMemoryError, even though the local variable data is no longer visible outside of the code block.So here comes the puzzle, that I'd like you to ponder a bit. If you very politely ask the VM to release memory, then you don't get an OutOfMemoryError: public class JavaMemoryPuzzlePolite {  private final int dataSize =      (int) (Runtime.getRuntime().maxMemory() * 0.6);  public void f() {    {      byte[] data = new byte[dataSize];    }    for(int i=0; i<10; i++) {      System.out.println("Please be so kind and release memory");    }    byte[] data2 = new byte[dataSize];  }  public static void main(String[] args) {    JavaMemoryPuzzlePolite jmp = new JavaMemoryPuzzlePolite();    jmp.f();    System.out.println("No OutOfMemoryError");  }}Why does this work? When I published this in my newsletter, I received over 400 emails from excited readers around the world, most of whom sent me the wrong explanation. After the 300th wrong answer, my replies became unfortunately a bit curt. Have a look at Issue 174 for a detailed explanation, but before you do, put on your thinking caps and try to figure it out yourself. Q: What do you think Java developers should know that they currently do not know?A: They should definitely get to know more about concurrency. It is a tough subject that most programmers try to avoid. Unfortunately we do come in contact with it. And when we do, we need to know how to protect ourselves and how to solve tricky system errors.Knowing your IDE is also useful. Most IDEs have a ton of shortcuts, which can make you a lot more productive in moving code around. Another thing that is useful is being able to read GC logs. Kirk Pepperdine has a great talk at JavaOne that I can recommend if you want to learn more. It's this: CON5405 – “Are Your Garbage Collection Logs Speaking to You?” Q: What are you looking forward to in Java 8?A: I'm quite excited about lambdas, though I must confess that I have not studied them in detail yet. Maurice Naftalin's Lambda FAQ is quite a good start to document what you can do with them. I'm looking forward to finding all the interesting bugs that we will now get due to lambdas obscuring what is really going on underneath, just like we had with generics.I am quite impressed with what the team at Oracle did with OpenJDK's performance. A lot of the benchmarks now run faster.Hopefully Java 8 will come with JSR 310, the Date and Time API. It still boggles my mind that such an important API has been left out in the cold for so long.What I am not looking forward to is losing perm space. Even though some systems run out of perm space, at least the problem is contained and they usually manage to work around it. In most cases, this is due to a memory leak in that region of memory. Once they bundle perm space with the old generation, I predict that memory leaks in perm space will be harder to find. More contracts for us, but also more pain for our customers.

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  • How to run boot loader in VMWare?

    - by Asim Haroon
    I am using Ubuntu as a virtual machine in VMWare. I have used this code to write a boot loader which would write Hello world on the screen. [BITS 16] [ORG 0x7C00] MOV SI, HelloString CALL PrintString JMP $ PrintCharacter: MOV AH, 0x0E MOV BH, 0x00 MOV BL, 0x07 INT 0x10 RET PrintString: next_character: MOV AL, [SI] INC SI OR AL, AL JZ exit_function CALL PrintCharacter JMP next_character exit_function: RET HelloString db 'Hello World', 0 TIMES 510 - ($ - $$) db 0 DW 0xAA55 I wrote this code in the text editor in Ubuntu and saved the file as Boot.asm Then I compiled the Boot.asm to boot.bin file by using this command nasm -f bin -o boot.bin Boot.asm and it didn't gave me any errors. After that I copied the boot.bin file to my usb and took it to my Windows OS. After this I burned the boot.bin file to boot.img and boot.iso files. Then I created a new virtual machine and named it booter, when it asked for the .iso file of the OS I want to run I gave it the boot.iso file, about which I told above, then I powered on that virtual machine but it gave me this error PXE-M0F: No boot filename received PXE-M0F: Exiting Intel PXE ROM Operating System not found Please tell me what is the main problem and how can I overcome that problem.

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  • MASM StrCmp Undefined?

    - by Yvan JANSSENS
    Hi, If I try to assemble the following code, I get a A2006 error ( error A2006: undefined symbol : StrCmp). Here's my code: .386 .model flat,stdcall option casemap:none include \masm32\include\windows.inc include \masm32\include\kernel32.inc include \masm32\include\masm32.inc include \masm32\include\user32.inc includelib \masm32\lib\kernel32.lib includelib \masm32\lib\masm32.lib includelib \masm32\lib\stdlib.lib includelib \masm32\lib\user32.lib .data YvanSoftware db "(c) YvanSoftware - ALL RIGHTS RESERVED", 13 ,10 ,0 EnterYourName db "Please enter your name: ", 0 CRLF db 13,10,0 TheHolyMan db "Yvan", 0 Seriously db "Seriously? You're the MAN!", 13,10,0 LoserName db "What a loser name.", 13,10 .data? buffer db 100 dup(?) .code start: invoke StdOut,addr YvanSoftware invoke StdOut, addr EnterYourName invoke StdIn, addr buffer, 100 invoke StdOut, addr CRLF invoke StrCmp,addr buffer, addr TheHolyMan ;error fires here je HolyMan IfNotHolyMan: invoke StdOut, addr LoserName jmp EndIfHolyMan HolyMan: invoke StdOut, addr Seriously jmp EndIfHolyMan EndIfHolyMan: invoke ExitProcess,0 END start I'm a complete n00b at assembler, and I'm trying to learn it. ;) Yvan

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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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  • 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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  • FASM vc MASM trasnlation problem in mov si, offset msg

    - by Ruben Trancoso
    hi folks, just did my first test with MASM and FASM with the same code (almos) and I falled in trouble. The only difference is that to produce just the 104 bytes I need to write to MBR in FASM I put org 7c00h and in MASM 0h. The problem is on the mov si, offset msg that in the first case transletes it to 44 7C (7c44h) and with masm translates to 44 00 (0044h)! but just when I change org 7c00h to org 0h in MASM. Otherwise it will produce the entire segment from 0 to 7dff. how do I solve it? or in short, how to make MASM produce a binary that begins at 7c00h as it first byte and subsequent jumps remain relative to 7c00h? .model TINY .code org 7c00h ; Boot entry point. Address 07c0:0000 on the computer memory xor ax, ax ; Zero out ax mov ds, ax ; Set data segment to base of RAM jmp start ; Jump to the first byte after DOS boot record data ; ---------------------------------------------------------------------- ; DOS boot record data ; ---------------------------------------------------------------------- brINT13Flag db 90h ; 0002h - 0EH for INT13 AH=42 READ brOEM db 'MSDOS5.0' ; 0003h - OEM name & DOS version (8 chars) brBPS dw 512 ; 000Bh - Bytes/sector brSPC db 1 ; 000Dh - Sectors/cluster brResCount dw 1 ; 000Eh - Reserved (boot) sectors brFATs db 2 ; 0010h - FAT copies brRootEntries dw 0E0h ; 0011h - Root directory entries brSectorCount dw 2880 ; 0013h - Sectors in volume, < 32MB brMedia db 240 ; 0015h - Media descriptor brSPF dw 9 ; 0016h - Sectors per FAT brSPH dw 18 ; 0018h - Sectors per track brHPC dw 2 ; 001Ah - Number of Heads brHidden dd 0 ; 001Ch - Hidden sectors brSectors dd 0 ; 0020h - Total number of sectors db 0 ; 0024h - Physical drive no. db 0 ; 0025h - Reserved (FAT32) db 29h ; 0026h - Extended boot record sig brSerialNum dd 404418EAh ; 0027h - Volume serial number (random) brLabel db 'OSAdventure' ; 002Bh - Volume label (11 chars) brFSID db 'FAT12 ' ; 0036h - File System ID (8 chars) ;------------------------------------------------------------------------ ; Boot code ; ---------------------------------------------------------------------- start: mov si, offset msg call showmsg hang: jmp hang msg db 'Loading...',0 showmsg: lodsb cmp al, 0 jz showmsgd push si mov bx, 0007 mov ah, 0eh int 10h pop si jmp showmsg showmsgd: retn ; ---------------------------------------------------------------------- ; Boot record signature ; ---------------------------------------------------------------------- dw 0AA55h ; Boot record signature END

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  • Help in building an 16 bit os

    - by Barshan Das
    I am trying to build an old 16 bit dos like os. My bootloader code: ; This is not my code. May be of Fritzos. I forgot the source. ORG 7c00h jmp Start drive db 0 msg db " Loader Initialization",0 msg2 db "ACos Loaded",0 print: lodsb cmp al, 0 je end mov ah, 0Eh int 10h jmp print end: ret Start: mov [ drive ], dl ; Get the floppy OS booted from ; Update the segment registers xor ax, ax ; XOR ax mov ds, ax ; Mov AX into DS mov si,msg call print ; Load Kernel. ResetFloppy: mov ax, 0x00 ; Select Floppy Reset BIOS Function mov dl, [ drive ] ; Select the floppy ADos booted from int 13h ; Reset the floppy drive jc ResetFloppy ; If there was a error, try again. ReadFloppy: mov bx, 0x9000 ; Load kernel at 9000h. mov ah, 0x02 ; Load disk data to ES:BX mov al, 17 ; Load two floppy head full's worth of data. mov ch, 0 ; First Cylinder mov cl, 2 ; Start at the 2nd Sector to load the Kernel mov dh, 0 ; Use first floppy head mov dl, [ drive ] ; Load from the drive kernel booted from. int 13h ; Read the floppy disk. jc ReadFloppy ; Error, try again. ; Clear text mode screen mov ax, 3 int 10h ;print starting message mov si,msg2 call print mov ax, 0x0 mov ss, ax mov sp, 0xFFFF jmp 9000h ; This part makes sure the bootsector is 512 bytes. times 510-($-$$) db 0 ;bootable sector signature dw 0xAA55 My example kernel code: asm(".code16\n"); void putchar(char); int main() { putchar('A'); return 0; } void putchar(char val) { asm("movb %0, %%al\n" "movb $0x0E, %%ah\n" "int $0x10\n" : :"r"(val) ) ; } This is how I compile it : nasm -f bin -o ./bin/boot.bin ./source/boot.asm gcc -nostdinc -fno-builtin -I./include -c -o ./bin/kernel.o ./source/kernel.c ld -Ttext=0x9000 -o ./bin/kernel.bin ./bin/kernel.o -e 0x0 dd if=/dev/zero of=./bin/empty.bin bs=1440K count=1 cat ./bin/boot.bin ./bin/kernel.bin ./bin/empty.bin|head -c 1440K > ./bin/os rm ./bin/empty.bin and I run it in virtual machine. When I make the putchar function ( in kernel code ) for constant value ....i.e like this: void putchar() { char val = 'A'; asm("movb %0, %%al\n" "movb $0x0E, %%ah\n" "int $0x10\n" : :"r"(val) ) ; } then it works fine. But when I pass argument to it ( That is in the previous code ) , then it prints a space for any character. What should I do?

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  • No Program Entry Point TASM Error

    - by Nathan Campos
    I'm trying to develop a simple kernel using TASM, using this code: ; beroset.asm ; ; This is a primitive operating system. ; ;********************************************************************** code segment para public use16 '_CODE' .386 assume cs:code, ds:code, es:code, ss:code org 0 Start: mov ax,cs mov ds,ax mov es,ax mov si,offset err_msg call DisplayMsg spin: jmp spin ;**************************************************************************** ; DisplayMsg ; ; displays the ASCIIZ message to the screen using int 10h calls ; ; Entry: ; ds:si ==> ASCII string ; ; Exit: ; ; Destroyed: ; none ; ; ;**************************************************************************** DisplayMsg proc push ax bx si cld nextchar: lodsb or al,al jz alldone mov bx,0007h mov ah,0eh int 10h jmp nextchar alldone: pop si bx ax ret DisplayMsg endp err_msg db "Operating system found and loaded.",0 code ends END Then I compile it like this: C:\DOCUME~1\Nathan\Desktop tasm /la /m2 beroset.asm Turbo Assembler Version 4.1 Copyright (c) 1988, 1996 Borland International Assembling file: beroset.asm Error messages: None Warning messages: None Passes: 2 Remaining memory: 406k C:\DOCUME~1\Nathan\Desktop tlink beroset, loader.bin Turbo Link Version 7.1.30.1. Copyright (c) 1987, 1996 Borland International Fatal: No program entry point C:\DOCUME~1\Nathan\Desktop What can I to correct this error?

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  • Intel IA-32 Assembly

    - by Kay
    I'm having a bit of difficulty converting the following java code into Intel IA-32 Assembly: class Person() { char name [8]; int age; void printName() {...} static void printAdults(Person [] list) { for(int k = 0; k < 100; k++){ if (list[k].age >= 18) { list[k].printName(); } } } } My attempt is: Person: push ebp; save callers ebp mov ebp, esp; setup new ebp push esi; esi will hold name push ebx; ebx will hold list push ecx; ecx will hold k init: mov esi, [ebp + 8]; mov ebx, [ebp + 12]; mov ecx, 0; k=0 forloop: cmp ecx, 100; jge end; if k>= 100 then break forloop cmp [ebx + 4 * ecx], 18 ; jl auxloop; if list[k].age < 18 then go to auxloop jmp printName; printName: auxloop: inc ecx; jmp forloop; end: pop ecx; pop ebx; pop esi; pop ebp; Is my code correct? NOTE: I'm not allowed to use global variables.

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  • Writing a VM - well formed bytecode?

    - by David Titarenco
    Hi, I'm writing a virtual machine in C just for fun. Lame, I know, but luckily I'm on SO so hopefully no one will make fun :) I wrote a really quick'n'dirty VM that reads lines of (my own) ASM and does stuff. Right now, I only have 3 instructions: add, jmp, end. All is well and it's actually pretty cool being able to feed lines (doing it something like write_line(&prog[1], "jmp", regA, regB, 0); and then running the program: while (machine.code_pointer <= BOUNDS && DONE != true) { run_line(&prog[machine.cp]); } I'm using an opcode lookup table (which may not be efficient but it's elegant) in C and everything seems to be working OK. My question is more of a "best practices" question but I do think there's a correct answer to it. I'm making the VM able to read binary files (storing bytes in unsigned char[]) and execute bytecode. My question is: is it the VM's job to make sure the bytecode is well formed or is it just the compiler's job to make sure the binary file it spits out is well formed? I only ask this because what would happen if someone would edit a binary file and screw stuff up (delete arbitrary parts of it, etc). Clearly, the program would be buggy and probably not functional. Is this even the VM's problem? I'm sure that people much smarter than me have figured out solutions to these problems, I'm just curious what they are!

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  • How do you update an Excel file (Data Refresh and update formulas) WITHOUT opening the file?

    - by Alex
    I have an Excel file that want to update and save automatically with out having to open it or manually interact with. Manually, I open the file up and hit data refresh which goes and does a SQL query and then hit F9 for the formulas to update and then I just close/save. (I then would mail the file out to people using a perl script or use SAS JMP to run some numbers/charts and also mail them out. Basically I need to script some things which require the XLS file to be updated.)

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  • "more than 3 levels of indentation, you're screwed" How should I understand this quote ?

    - by jokoon
    The answer to that is that if you need more than 3 levels of indentation, you're screwed anyway, and should fix your program. What can I deduct from this quote ? On top of the fact that too long methods are hard to maintain, are they hard or impossible to optimize for the compiler ? I don't really understand if this quote encourages better coding practice or is really a mathematical/algorithmic sort of truth... I also read in some C++ optimizing guide that dividing up a program into more function improves its design is a common thing taught at school, but it should be not done too much, since it can turn into a lot of JMP calls (even if the compiler can inline some methods by itself).

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  • If you need more than 3 levels of indentation, you're screwed?

    - by jokoon
    Per the Linux kernel coding style document: The answer to that is that if you need more than 3 levels of indentation, you're screwed anyway, and should fix your program. What can I deduct from this quote? On top of the fact that too long methods are hard to maintain, are they hard or impossible to optimize for the compiler? I don't really understand if this quote encourages better coding practice or is really a mathematical / algorithmic sort of truth. I also read in some C++ optimizing guide that dividing up a program into more function improves its design is a common thing taught at school, but it should be not done too much, since it can turn into a lot of JMP calls (even if the compiler can inline some methods by itself).

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  • If you need more than 3 levels of indentation, you're screwed?

    - by jokoon
    Per the Linux kernel coding style document: The answer to that is that if you need more than 3 levels of indentation, you're screwed anyway, and should fix your program. What can I deduce from this quote? On top of the fact that too long methods are hard to maintain, are they hard or impossible to optimize for the compiler? I don't really understand if this quote encourages better coding practice or is really a mathematical / algorithmic sort of truth. I also read in some C++ optimizing guide that "dividing up a program into more functions improves its design" is frequently taught in CS courses, but it should be not done too much, since it can turn into a lot of JMP calls (even if the compiler can inline some methods by itself).

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  • How is printf() implemented in c?

    - by Mask
    Disassembling printf doesn't give much info: (gdb) disas printf Dump of assembler code for function printf: 0x00401b38 <printf+0>: jmp *0x405130 0x00401b3e <printf+6>: nop 0x00401b3f <printf+7>: nop End of assembler dump. How is it implemented under the hood? Why disassembling doesn't help? What does * mean before 0x405130?

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  • What does * address(found in printf) mean in assembly?

    - by Mask
    Disassembling printf doesn't give much info: (gdb) disas printf Dump of assembler code for function printf: 0x00401b38 <printf+0>: jmp *0x405130 0x00401b3e <printf+6>: nop 0x00401b3f <printf+7>: nop End of assembler dump. (gdb) disas 0x405130 Dump of assembler code for function _imp__printf: 0x00405130 <_imp__printf+0>: je 0x405184 <_imp__vfprintf+76> 0x00405132 <_imp__printf+2>: add %al,(%eax) How is it implemented under the hood? Why disassembling doesn't help? What does * mean before 0x405130?

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  • How is prinf() implemented in c?

    - by Mask
    Disassembling printf doesn't give much info: (gdb) disas printf Dump of assembler code for function printf: 0x00401b38 <printf+0>: jmp *0x405130 0x00401b3e <printf+6>: nop 0x00401b3f <printf+7>: nop End of assembler dump. How is it implemented under the hood?

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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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  • variables in assembler

    - by stupid_idiot
    hi, i know this is kinda retarded but I just can't figure it out. I'm debugging this: xor eax,eax mov ah,[var1] mov al,[var2] call addition stop: jmp stop var1: db 5 var2: db 6 addition: add ah,al ret the numbers that I find on addresses var1 and var2 are 0x0E and 0x07. I know it's not segmented, but that ain't reason for it to do such escapades, because the addition call works just fine. Could you please explain to me where is my mistake?

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  • How to obtain a pointer out of a C++ vtable?

    - by Josh Haberman
    Say you have a C++ class like: class Foo { public: virtual ~Foo() {} virtual DoSomething() = 0; }; The C++ compiler translates a call into a vtable lookup: Foo* foo; // Translated by C++ to: // foo->vtable->DoSomething(foo); foo->DoSomething(); Suppose I was writing a JIT compiler and I wanted to obtain the address of the DoSomething() function for a particular instance of class Foo, so I can generate code that jumps to it directly instead of doing a table lookup and an indirect branch. My questions are: Is there any standard C++ way to do this (I'm almost sure the answer is no, but wanted to ask for the sake of completeness). Is there any remotely compiler-independent way of doing this, like a library someone has implemented that provides an API for accessing a vtable? I'm open to completely hacks, if they will work. For example, if I created my own derived class and could determine the address of its DoSomething method, I could assume that the vtable is the first (hidden) member of Foo and search through its vtable until I find my pointer value. However, I don't know a way of getting this address: if I write &DerivedFoo::DoSomething I get a pointer-to-member, which is something totally different. Maybe I could turn the pointer-to-member into the vtable offset. When I compile the following: class Foo { public: virtual ~Foo() {} virtual void DoSomething() = 0; }; void foo(Foo *f, void (Foo::*member)()) { (f->*member)(); } On GCC/x86-64, I get this assembly output: Disassembly of section .text: 0000000000000000 <_Z3fooP3FooMS_FvvE>: 0: 40 f6 c6 01 test sil,0x1 4: 48 89 74 24 e8 mov QWORD PTR [rsp-0x18],rsi 9: 48 89 54 24 f0 mov QWORD PTR [rsp-0x10],rdx e: 74 10 je 20 <_Z3fooP3FooMS_FvvE+0x20> 10: 48 01 d7 add rdi,rdx 13: 48 8b 07 mov rax,QWORD PTR [rdi] 16: 48 8b 74 30 ff mov rsi,QWORD PTR [rax+rsi*1-0x1] 1b: ff e6 jmp rsi 1d: 0f 1f 00 nop DWORD PTR [rax] 20: 48 01 d7 add rdi,rdx 23: ff e6 jmp rsi I don't fully understand what's going on here, but if I could reverse-engineer this or use an ABI spec I could generate a fragment like the above for each separate platform, as a way of obtaining a pointer out of a vtable.

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  • Web Client Service constantly in 'Stopping' state

    - by Mark
    I have a user who's Web Client service constantly reports that it's in the 'Stopping' state and it's hindering her ability to save JMP files to a SharePoint site using the UNC path. She's running Windows XP Service Pack 3. I've tried modifying the Web Client parameters in the registry for UseBasicAuth and FileAttributesLimitInBytes with no luck. When I set the service to Manual and then try to start it after Windows boots up, it starts and then immediately goes into the Stopping state again. Other things I've tried: Removing/Reinstalling her network card Removing/Reinstalling the Client for Microsoft Networks and File and Print Sharing Checked that the BITS and RPC services are running fine (not sure if they're related) Does anyone have any other ideas? Is there a way to repair/rebuild the Web Client service?

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