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  • Code Golf: Collatz Conjecture

    - by Earlz
    Inspired by http://xkcd.com/710/ here is a code golf for it. The Challenge Given a positive integer greater than 0, print out the hailstone sequence for that number. The Hailstone Sequence See Wikipedia for more detail.. If the number is even, divide it by two. If the number is odd, triple it and add one. Repeat this with the number produced until it reaches 1. (if it continues after 1, it will go in an infinite loop of 1 -> 4 -> 2 -> 1...) Sometimes code is the best way to explain, so here is some from Wikipedia function collatz(n) show n if n > 1 if n is odd call collatz(3n + 1) else call collatz(n / 2) This code works, but I am adding on an extra challenge. The program must not be vulnerable to stack overflows. So it must either use iteration or tail recursion. Also, bonus points for if it can calculate big numbers and the language does not already have it implemented. (or if you reimplement big number support using fixed-length integers) Test case Number: 21 Results: 21 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1 Number: 3 Results: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 Also, the code golf must include full user input and output.

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  • Getting Error System.Runtime.InteropServices.COMException

    - by Savan Parmar
    Hey All, I am using Vb.Net to Create Labels in Microsoft. For that i am using below mantioend Code. Public Sub CreateLabel(ByVal StrFilter As String, ByVal Path As String) WordApp = CreateObject("Word.Application") ''Add a new document. WordDoc = WordApp.Documents.Add() Dim oConn As SqlConnection = New SqlConnection(connSTR) oConn.Open() Dim oCmd As SqlCommand Dim oDR As SqlDataReader oCmd = New SqlCommand(StrFilter, oConn) oDR = oCmd.ExecuteReader Dim intI As Integer Dim FilePath As String = "" With WordDoc.MailMerge With .Fields Do While oDR.Read For intI = 0 To oDR.FieldCount - 1 .Add(WordApp.Selection.Range, oDR.Item(intI)) Next Loop End With Dim objAutoText As Word.AutoTextEntry = WordApp.NormalTemplate.AutoTextEntries.Add("MyLabelLayout", WordDoc.Content) WordDoc.Content.Delete() .MainDocumentType = Word.WdMailMergeMainDocType.wdMailingLabels FilePath = CreateSource(StrFilter) .OpenDataSource(FilePath) Dim NewLabel As Word.CustomLabel = WordApp.MailingLabel.CustomLabels.Add("MyLabel", False) WordApp.MailingLabel.CreateNewDocument(Name:="MyLabel", Address:="", AutoText:="MyLabelLayout") objAutoText.Delete() .Destination = Word.WdMailMergeDestination.wdSendToNewDocument WordApp.Visible = True .Execute() End With oConn.Close() WordDoc.Close() End Sub Private Function CreateSource(ByVal StrFilter As String) As String Dim CnnUser As SqlConnection = New SqlConnection(connSTR) Dim sw As StreamWriter = File.CreateText("C:\Mail.Txt") Dim Path As String = "C:\Mail.Txt" Dim StrHeader As String = "" Try Dim SelectCMD As SqlCommand = New SqlCommand(StrFilter, CnnUser) Dim oDR As SqlDataReader Dim IntI As Integer SelectCMD.CommandType = CommandType.Text CnnUser.Open() oDR = SelectCMD.ExecuteReader For IntI = 0 To oDR.FieldCount - 1 StrHeader &= oDR.GetName(IntI) & " ," Next StrHeader = Mid(StrHeader, 1, Len(StrHeader) - 2) sw.WriteLine(StrHeader) sw.Flush() sw.Close() StrHeader = "" Do While oDR.Read For IntJ As Integer = 0 To oDR.FieldCount - 1 StrHeader &= oDR.GetString(IntJ) & " ," Next Loop StrHeader = Mid(StrHeader, 1, Len(StrHeader) - 2) sw = File.AppendText(Path) sw.WriteLine(StrHeader) CnnUser.Close() sw.Flush() sw.Close() Catch ex As Exception MessageBox.Show(ex.Message, "TempID", MessageBoxButtons.OK, MessageBoxIcon.Error) End Try Return Path End Function Now when i am running the programm i am getting this error.I tried hard but not able to locate what could be the problem the error is:- System.Runtime.InteropServices.COMException --Horizontal and vertical pitch must be greater than or equal to the label width and height, respectively. Even though i tried to set the Horizontal and vertical pitch programatically but it gives same err. Plz if any one can help

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  • Delphi Phrase Count / Keyword Density

    - by Brad
    Does anyone know how to or have some code on counting the number of unique phrases in a document? (Single word, two word phrases, three word phrases). Thanks Example of what I'm looking for: What I mean is I have a text document, and i need to see what the most popular word phrases are. Example text I took the car to the car wash. I : 1 took : 1 the : 2 car: 2 to : 1 wash : 1 I took : 1 took the : 1 the car : 2 car to : 1 to the : 1 car wash : 1 I took the : 1 took the car : 1 the car to : 1 car to the : 1 to the car : 1 the car wash : 1 I took the car to : 1 took the car to the : 1 the car to the car : 1 car to the car wash : 1 I need the phrase, and the count that it shows up. Any help would be appreciated. The closet thing I found to this was a PHP script from http://tools.seobook.com/general/keyword-density/source.php I used to have some code for this, but I cannot find it.

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  • Increment sequence when updating freebusy in icalendar?

    - by user302038
    I'm new to icalendar, but I couldn't find an answer to this specific situation. My web site has a shared calendar where users can add and update events. They can also indicate on the web site if they will be attending an event. I want to add a button to let them download the calendar as published icalendar events. When they indicated they will be attending an event, I want the event in the icalendar file to be set as "busy", otherwise as "not busy". So I can easily increment the sequence number on the event if the event is changed, but must I increment the sequence number if the user changed whether or not they will attend the event? I don't think that is a "mandatory" reason to increment the sequence, but I'm not sure. What will happen if the user changes their busy status for an event and re-downloads without the sequence number changing? I suppose I can keep a separate sequence number for each time the user changes their attendance status, then add the event sequence number and attendance sequence number together for the "final" event sequence number that's downloaded, but do I have to do it that way?

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  • Problem calling linux C code from FIQ handler

    - by fastmonkeywheels
    I'm working on an armv6 core and have an FIQ hander that works great when I do all of my work in it. However I need to branch to some additional code that's too large for the FIQ memory area. The FIQ handler gets copied from fiq_start to fiq_end to 0xFFFF001C when registered static void test_fiq_handler(void) { asm volatile("\ .global fiq_start\n\ fiq_start:"); // clear gpio irq asm("ldr r10, GPIO_BASE_ISR"); asm("ldr r9, [r10]"); asm("orr r9, #0x04"); asm("str r9, [r10]"); // clear force register asm("ldr r10, AVIC_BASE_INTFRCH"); asm("ldr r9, [r10]"); asm("mov r9, #0"); asm("str r9, [r10]"); // prepare branch register asm(" ldr r11, fiq_handler"); // save all registers, build sp and branch to C asm(" adr r9, regpool"); asm(" stmia r9, {r0 - r8, r14}"); asm(" adr sp, fiq_sp"); asm(" ldr sp, [sp]"); asm(" add lr, pc,#4"); asm(" mov pc, r11"); #if 0 asm("ldr r10, IOMUX_ADDR12"); asm("ldr r9, [r10]"); asm("orr r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); asm("bic r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); #endif asm(" adr r9, regpool"); asm(" ldmia r9, {r0 - r8, r14}"); // return asm("subs pc, r14, #4"); asm("IOMUX_ADDR12: .word 0xFC2A4000"); asm("AVIC_BASE_INTCNTL: .word 0xFC400000"); asm("AVIC_BASE_INTENNUM: .word 0xFC400008"); asm("AVIC_BASE_INTDISNUM: .word 0xFC40000C"); asm("AVIC_BASE_FIVECSR: .word 0xFC400044"); asm("AVIC_BASE_INTFRCH: .word 0xFC400050"); asm("GPIO_BASE_ISR: .word 0xFC2CC018"); asm(".globl fiq_handler"); asm("fiq_sp: .long fiq_stack+120"); asm("fiq_handler: .long 0"); asm("regpool: .space 40"); asm(".pool"); asm(".align 5"); asm("fiq_stack: .space 124"); asm(".global fiq_end"); asm("fiq_end:"); } fiq_hander gets set to the following function: static void fiq_flip_pins(void) { asm("ldr r10, IOMUX_ADDR12_k"); asm("ldr r9, [r10]"); asm("orr r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); asm("bic r9, #0x08 @ top/vertex LED"); asm("str r9,[r10] @turn on LED"); asm("IOMUX_ADDR12_k: .word 0xFC2A4000"); } EXPORT_SYMBOL(fiq_flip_pins); I know that since the FIQ handler operates outside of any normal kernel API's and that it is a rather high priority interrupt I must ensure that whatever I call is already swapped into memory. I do this by having the fiq_flip_pins function defined in the monolithic kernel and not as a module which gets vmalloc. If I don't branch to the fiq_flip_pins function, and instead do the work in the test_fiq_handler function everything works as expected. It's the branching that's causing me problems at the moment. Right after branching I get a kernel panic about a paging request. I don't understand why I'm getting the paging request. fiq_flip_pins is in the kernel at: c00307ec t fiq_flip_pins Unable to handle kernel paging request at virtual address 736e6f63 pgd = c3dd0000 [736e6f63] *pgd=00000000 Internal error: Oops: 5 [#1] PREEMPT Modules linked in: hello_1 CPU: 0 Not tainted (2.6.31-207-g7286c01-svn4 #122) PC is at strnlen+0x10/0x28 LR is at string+0x38/0xcc pc : [<c016b004>] lr : [<c016c754>] psr: a00001d3 sp : c3817ea0 ip : 736e6f63 fp : 00000400 r10: c03cab5c r9 : c0339ae0 r8 : 736e6f63 r7 : c03caf5c r6 : c03cab6b r5 : ffffffff r4 : 00000000 r3 : 00000004 r2 : 00000000 r1 : ffffffff r0 : 736e6f63 Flags: NzCv IRQs off FIQs off Mode SVC_32 ISA ARM Segment user Control: 00c5387d Table: 83dd0008 DAC: 00000015 Process sh (pid: 1663, stack limit = 0xc3816268) Stack: (0xc3817ea0 to 0xc3818000) Since there are no API calls in my code I have to assume that something is going wrong in the C call and back. Any help solving this is appreciated.

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  • Java: If vs. Switch

    - by _ande_turner_
    I have a piece of code with a) which I replaced with b) purely for legibility ... a) if ( WORD[ INDEX ] == 'A' ) branch = BRANCH.A; /* B through to Y */ if ( WORD[ INDEX ] == 'Z' ) branch = BRANCH.Z; b) switch ( WORD[ INDEX ] ) { case 'A' : branch = BRANCH.A; break; /* B through to Y */ case 'Z' : branch = BRANCH.Z; break; } ... will the switch version cascade through all the permutations or jump to a case ? EDIT: Some of the answers below regard alternative approaches to the approach above. I have included the following to provide context for its use. The reason I asked, the Question above, was because the speed of adding words empirically improved. This isn't production code by any means, and was hacked together quickly as a PoC. The following seems to be a confirmation of failure for a thought experiment. I may need a much bigger corpus of words than the one I am currently using though. The failure arises from the fact I did not account for the null references still requiring memory. ( doh ! ) public class Dictionary { private static Dictionary ROOT; private boolean terminus; private Dictionary A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z; private static Dictionary instantiate( final Dictionary DICTIONARY ) { return ( DICTIONARY == null ) ? new Dictionary() : DICTIONARY; } private Dictionary() { this.terminus = false; this.A = this.B = this.C = this.D = this.E = this.F = this.G = this.H = this.I = this.J = this.K = this.L = this.M = this.N = this.O = this.P = this.Q = this.R = this.S = this.T = this.U = this.V = this.W = this.X = this.Y = this.Z = null; } public static void add( final String...STRINGS ) { Dictionary.ROOT = Dictionary.instantiate( Dictionary.ROOT ); for ( final String STRING : STRINGS ) Dictionary.add( STRING.toUpperCase().toCharArray(), Dictionary.ROOT , 0, STRING.length() - 1 ); } private static void add( final char[] WORD, final Dictionary BRANCH, final int INDEX, final int INDEX_LIMIT ) { Dictionary branch = null; switch ( WORD[ INDEX ] ) { case 'A' : branch = BRANCH.A = Dictionary.instantiate( BRANCH.A ); break; case 'B' : branch = BRANCH.B = Dictionary.instantiate( BRANCH.B ); break; case 'C' : branch = BRANCH.C = Dictionary.instantiate( BRANCH.C ); break; case 'D' : branch = BRANCH.D = Dictionary.instantiate( BRANCH.D ); break; case 'E' : branch = BRANCH.E = Dictionary.instantiate( BRANCH.E ); break; case 'F' : branch = BRANCH.F = Dictionary.instantiate( BRANCH.F ); break; case 'G' : branch = BRANCH.G = Dictionary.instantiate( BRANCH.G ); break; case 'H' : branch = BRANCH.H = Dictionary.instantiate( BRANCH.H ); break; case 'I' : branch = BRANCH.I = Dictionary.instantiate( BRANCH.I ); break; case 'J' : branch = BRANCH.J = Dictionary.instantiate( BRANCH.J ); break; case 'K' : branch = BRANCH.K = Dictionary.instantiate( BRANCH.K ); break; case 'L' : branch = BRANCH.L = Dictionary.instantiate( BRANCH.L ); break; case 'M' : branch = BRANCH.M = Dictionary.instantiate( BRANCH.M ); break; case 'N' : branch = BRANCH.N = Dictionary.instantiate( BRANCH.N ); break; case 'O' : branch = BRANCH.O = Dictionary.instantiate( BRANCH.O ); break; case 'P' : branch = BRANCH.P = Dictionary.instantiate( BRANCH.P ); break; case 'Q' : branch = BRANCH.Q = Dictionary.instantiate( BRANCH.Q ); break; case 'R' : branch = BRANCH.R = Dictionary.instantiate( BRANCH.R ); break; case 'S' : branch = BRANCH.S = Dictionary.instantiate( BRANCH.S ); break; case 'T' : branch = BRANCH.T = Dictionary.instantiate( BRANCH.T ); break; case 'U' : branch = BRANCH.U = Dictionary.instantiate( BRANCH.U ); break; case 'V' : branch = BRANCH.V = Dictionary.instantiate( BRANCH.V ); break; case 'W' : branch = BRANCH.W = Dictionary.instantiate( BRANCH.W ); break; case 'X' : branch = BRANCH.X = Dictionary.instantiate( BRANCH.X ); break; case 'Y' : branch = BRANCH.Y = Dictionary.instantiate( BRANCH.Y ); break; case 'Z' : branch = BRANCH.Z = Dictionary.instantiate( BRANCH.Z ); break; } if ( INDEX == INDEX_LIMIT ) branch.terminus = true; else Dictionary.add( WORD, branch, INDEX + 1, INDEX_LIMIT ); } public static boolean is( final String STRING ) { Dictionary.ROOT = Dictionary.instantiate( Dictionary.ROOT ); return Dictionary.is( STRING.toUpperCase().toCharArray(), Dictionary.ROOT, 0, STRING.length() - 1 ); } private static boolean is( final char[] WORD, final Dictionary BRANCH, final int INDEX, final int INDEX_LIMIT ) { Dictionary branch = null; switch ( WORD[ INDEX ] ) { case 'A' : branch = BRANCH.A; break; case 'B' : branch = BRANCH.B; break; case 'C' : branch = BRANCH.C; break; case 'D' : branch = BRANCH.D; break; case 'E' : branch = BRANCH.E; break; case 'F' : branch = BRANCH.F; break; case 'G' : branch = BRANCH.G; break; case 'H' : branch = BRANCH.H; break; case 'I' : branch = BRANCH.I; break; case 'J' : branch = BRANCH.J; break; case 'K' : branch = BRANCH.K; break; case 'L' : branch = BRANCH.L; break; case 'M' : branch = BRANCH.M; break; case 'N' : branch = BRANCH.N; break; case 'O' : branch = BRANCH.O; break; case 'P' : branch = BRANCH.P; break; case 'Q' : branch = BRANCH.Q; break; case 'R' : branch = BRANCH.R; break; case 'S' : branch = BRANCH.S; break; case 'T' : branch = BRANCH.T; break; case 'U' : branch = BRANCH.U; break; case 'V' : branch = BRANCH.V; break; case 'W' : branch = BRANCH.W; break; case 'X' : branch = BRANCH.X; break; case 'Y' : branch = BRANCH.Y; break; case 'Z' : branch = BRANCH.Z; break; } if ( branch == null ) return false; if ( INDEX == INDEX_LIMIT ) return branch.terminus; else return Dictionary.is( WORD, branch, INDEX + 1, INDEX_LIMIT ); } }

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  • Link List Problem,

    - by david
    OK i have a problem with a Link List program i'm trying to Do, the link List is working fine. Here is my code #include <iostream> using namespace std; struct record { string word; struct record * link; }; typedef struct record node; node * insert_Node( node * head, node * previous, string key ); node * search( node *head, string key, int *found); void displayList(node *head); node * delete_node( node *head, node * previous, string key); int main() { node * previous, * head = NULL; int found = 0; string node1Data,newNodeData, nextData,lastData; //set up first node cout <<"Depature"<<endl; cin >>node1Data; previous = search( head, node1Data, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, node1Data); cout <<"Depature inserted"<<endl; //insert node between first node and head cout <<"Destination"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Destinationinserted"<<endl; //insert node between second node and head cout <<"Cost"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Cost inserted"<<endl; cout <<"Number of Seats Required"<<endl; //place node between new node and first node cin >>nextData; previous = search( head, nextData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, nextData); cout <<"Number of Seats Required inserted"<<endl; //insert node between first node and head cout <<"Name"<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Name inserted"<<endl; //insert node between node and head cout <<"Address "<<endl; cin >>newNodeData; previous = search( head, newNodeData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, newNodeData); cout <<"Address inserted"<<endl; //place node as very last node cin >>lastData; previous = search( head, lastData, &found); cout <<"Previous" <<previous<<endl; head = insert_Node(head, previous, lastData); cout <<"C"<<endl; displayList(head); char Ans = 'y'; //Delete nodes do { cout <<"Enter Keyword to be delete"<<endl; cin >>nextData; previous = search( head, nextData, &found); if (found == 1) head = delete_node( head, previous,nextData); displayList(head); cout <<"Do you want to Delete more y /n "<<endl; cin >> Ans; } while( Ans =='y'); int choice, i=0, counter=0; int fclass[10]; int coach[10]; printf("Welcome to the booking program"); printf("\n-----------------"); do{ printf("\n Please pick one of the following option:"); printf("\n 1) Reserve a first class seat on Flight 101."); printf("\n 2) Reserve a coach seat on Flight 101."); printf("\n 3) Quit "); printf("\n ---------------------------------------------------------------------"); printf("\nYour choice?"); scanf("%d",&choice); switch(choice) { case 1: i++; if (i <10){ printf("Here is your seat: %d " , fclass[i]); } else if (i = 10) { printf("Sorry there is no more seats on First Class. Please wait for the next flight"); } break; case 2: if (i <10){ printf("Here is your Seat Coach: %d " , coach[i]); } else if ( i = 10) { printf("Sorry their is no more Seats on Coach. Please wait for the next flight"); } break; case 3: printf("Thank you and goodbye\n"); //exit(0); } } while (choice != 3); } /******************************************************* search function to return previous position of node ******************************************************/ node * search( node *head, string key, int *found) { node * previous, * current; current = head; previous = current; *found = 0;//not found //if (current->word < key) move through links until the next link //matches or current_word > key while( current !=NULL) { //compare exactly if (key ==current->word ) { *found = 1; break; } //if key is less than word else if ( key < current->word ) break; else { //previous stays one link behind current previous = current; current = previous -> link; } } return previous; } /******************************************************** display function as used with createList ******************************************************/ void displayList(node *head) { node * current; //current now contains the address held of the 1st node similar //to head current = head; cout << "\n\n"; if( current ==NULL) cout << "Empty List\n\n"; else { /*Keep going displaying the contents of the list and set current to the address of the next node. When set to null, there are no more nodes */ while(current !=NULL) { cout << current->word<<endl; current = current ->link; } } } /************************************************************ insert node used to position node (i) empty list head = NULL (ii) to position node before the first node key < head->word (iii) every other position including the end of the list This is done using the following steps (a) Pass in all the details to create the node either details or a whole record (b) Pass the details over to fill the node (C) Use the if statement to add the node to the list **********************************************************/ node * insert_Node( node * head, node * previous, string key ) { node * new_node, * temp; new_node = new node; //create the node new_node ->word = key; new_node -> link = NULL; if (head == NULL || key < head->word ) //empty list { //give address of head to temp temp = head; //head now points to the new_node head = new_node; //new_node now points to what head was pointing at new_node -> link = temp; } else { //pass address held in link to temp temp = previous-> link; //put address of new node to link of previous previous -> link = new_node; //pass address of temp to link of new node new_node -> link = temp; } return head; } node * delete_node( node *head, node * previous, string key) { /* this function will delete a node but will not return its contents */ node * temp; if(key == head->word) //delete node at head of list { temp = head; //point head at the next node head = head -> link; } else { //holds the address of the node after the one // to be deleted temp = previous-> link; /*assign the previous to the address of the present node to be deleted which holds the address of the next node */ previous-> link = previous-> link-> link; } delete temp; return head; }//end delete The problem i have is when i Enter in the Number 2 in the Node(Seats) i like to get a Counter Taken 2 off of 50, some thing like what i have here enter code here int choice, i=0, counter=0; int fclass[10]; int coach[10]; printf("Welcome to the booking program"); printf("\n-----------------"); do{ printf("\n Please pick one of the following option:"); printf("\n 1) Reserve a first class seat on Flight 101."); printf("\n 2) Reserve a coach seat on Flight 101."); printf("\n 3) Quit "); printf("\n ---------------------------------------------------------------------"); printf("\nYour choice?"); scanf("%d",&choice); switch(choice) { case 1: i++; if (i <10){ printf("Here is your seat: %d " , fclass[i]); } else if (i = 10) { printf("Sorry there is no more seats on First Class. Please wait for the next flight"); } break; case 2: if (i <10){ printf("Here is your Seat Coach: %d " , coach[i]); } else if ( i = 10) { printf("Sorry their is no more Seats on Coach. Please wait for the next flight"); } break; case 3: printf("Thank you and goodbye\n"); //exit(0); } } while (choice != 3); How can i get what the User enters into number of Seats into this function

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  • Newbie Python programmer tangling with Lists.

    - by Sergio Tapia
    Here's what I've got so far: # A. match_ends # Given a list of strings, return the count of the number of # strings where the string length is 2 or more and the first # and last chars of the string are the same. # Note: python does not have a ++ operator, but += works. def match_ends(words): counter = 0 for word in words: if len(word) >= 2 and word[0] == word[-1]: counter += counter return counter # +++your code here+++ return I'm following the Google Python Class, so this isn't homework, but me just learning and improving myself; so please no negative comments about 'not doing my homework'. :P What do you guys think I'm doing wrong here? Here's the result: match_ends X got: 0 expected: 3 X got: 0 expected: 2 X got: 0 expected: 1 I'm really loving Python, so I just know that I'll get better at it. :)

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  • Delay On Assembler?

    - by Norm
    Hey, I want to know how i can do delay (Timer) on assembler 16 bit on PC. Thank You for helping, Norm. OS: Windows CODE: delay: inc bx cmp bx,WORD ptr[time] je delay2 jmp delay delay2: inc dx cmp dx,WORD ptr[time2] je delay3 jmp delay mov bx,0 delay3: inc cx cmp cx,WORD ptr[time3] je Finish_delay jmp delay its not work good i need less complicated code

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  • Serial Communication between Java RXTX and Arduino

    - by SharpBarb
    I'm trying to communicate between my PC (Windows 7 using Netbeans and RXTX) with an Arduino Pro, using the serial port. The Arduino is actually connected to the PC using an FTDI cable. The code is based on the Java SimpleRead.Java found here. Currently the Arduino simply prints out a string when it starts up. My Java program should print the number of bytes that have been read and then print out the contents. The Java program works, sort of... If the string is long (10 bytes or so) the output will get broken up. So if on the Arduino I print Serial.println("123456789123456789"); //20 bytes including '\r' and '\n' The output of my Java program may look something like: Number of Bytes: 15 1234567891234 Number of Bytes: 5 56789 or Number of Bytes: 12 1234567891 Number of Bytes: 8 23456789 I'm thinking it's a timing problem, because when I manually go through the code using the debugger, the result string is always what it should be: one 20 byte string. I've been messing with various things but I haven't been able to fix the problem. Here is the part of the code that is giving me problems: static int baudrate = 9600, dataBits = SerialPort.DATABITS_8, stopBits = SerialPort.STOPBITS_1, parity = SerialPort.PARITY_NONE; byte[] readBuffer = new byte[128]; ... ... public void serialEvent(SerialPortEvent event) { if (event.getEventType() == SerialPortEvent.DATA_AVAILABLE) { try { if (input.available() > 0) { //Read the InputStream and return the number of bytes read numBytes = input.read(readBuffer); String result = new String(readBuffer,0,numBytes); System.out.println("Number of Bytes: " + numBytes); System.out.println(result); } } catch (IOException e) { System.out.println("Data Available Exception"); } }

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  • What Algorithm will Find New Longtail Keywords for *keyword* in PPC

    - by Becci
    I am looking for the algorithm (or combo) that would allow someone to find new longtail PPC search phrases based on say one corekeyword. Eg #1 word word corekeyword eg #2 word corekeyword word Google search tool allows a limited number vertically - mostly of eg#1 (https://adwords.google.com.au/select/KeywordToolExternal) I also know of other PPC apps that allow more volume than google adwords keyword tool, But I want to find other combos that mention the corekeyword & then naturally sort for the highest volume searched. Working example of exact match: corekeyword: copywriter (40,500 searches a month) google will serve up: become a copywriter (480 searches globally/month in english) But if I specifically look up: How to become a copywriter (720 searches a month) This exact longtail keyword phrase has 300 more searches than the 3 word version spat out by google. I want the algorithm to find any other highly search exact longtials like: how to become a copywriter Simply because it was save significant $ finding other longtail keywords after your campaign has been running an made google lots of money. I don't want a concantenation algorithm (I already have one of those), because hypothetically, I don't know what keywords will be that I want to find. Any gurus out there? Becci

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  • AS3 Regular Expression Question...

    - by Coltech
    Can someone give me a regular expression that will verify if all the letters in the word "cat" were also in the word "coating" in the proper sequence? So for the word "coating", the RegEx will test true for "cat" but false for "act".

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  • Delphi Phrase Count

    - by Brad
    Does anyone know how to or have some code on counting the number of unique phrases in a document? (Single word, two word phrases, three word phrases). Thanks

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  • Xcode & passing command line arguments

    - by Brisco
    I just started working with C & Xcode and I've run into a little difficulty. All I want to do is read a file from the command line and see the output in the terminal. I think my problem lies with the path to the file that I want to read in. I'm using a Mac and the file is on my desktop, so the path should be Users/myName/Desktop/words.txt. Is this correct? This is my code: #import <Foundation/Foundation.h> int main (int argc, const char* argv[]){ if(argc == 1){ NSLog(@" you must pass at least one arguement"); return 1; } NSLog(@"russ"); FILE* wordFile = fopen(argv[1] , "r"); char word[100]; while (fgets(word,100,wordFile)) { NSLog(@" %s is %d chars long", word,strlen(word)); } fclose(wordFile); return 0; }//main

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  • Big-O of PHP functions?

    - by Kendall Hopkins
    After using PHP for a while now, I've noticed that not all PHP built in functions as fast as expected. Consider the below two possible implementations of a function that finds if a number is prime using a cached array of primes. //very slow for large $prime_array $prime_array = array( 2, 3, 5, 7, 11, 13, .... 104729, ... ); $result_array = array(); foreach( $array_of_number => $number ) { $result_array[$number] = in_array( $number, $large_prime_array ); } //still decent performance for large $prime_array $prime_array => array( 2 => NULL, 3 => NULL, 5 => NULL, 7 => NULL, 11 => NULL, 13 => NULL, .... 104729 => NULL, ... ); foreach( $array_of_number => $number ) { $result_array[$number] = array_key_exists( $number, $large_prime_array ); } This is because in_array is implemented with a linear search O(n) which will linearly slow down as $prime_array grows. Where the array_key_exists function is implemented with a hash lookup O(1) which will not slow down unless the hash table gets extremely populated (in which case it's only O(logn)). So far I've had to discover the big-O's via trial and error, and occasionally looking at the source code. Now for the question... I was wondering if there was a list of the theoretical (or practical) big O times for all* the PHP built in functions. *or at least the interesting ones For example find it very hard to predict what the big O of functions listed because the possible implementation depends on unknown core data structures of PHP: array_merge, array_merge_recursive, array_reverse, array_intersect, array_combine, str_replace (with array inputs), etc.

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  • Order Result in Sqlite

    - by saturngod
    In MySQL , my sql is like following SELECT * , IF( `Word` = 'sim', 1, IF( `Word` LIKE 'sim%', 2, IF( `Word` LIKE '%sim', 4, 3 ) ) ) AS `sort` FROM `dblist` WHERE `Word` LIKE '%sim%' ORDER BY `sort` , `Word` This sql is not working in SQlite. I want to do result order. SELECT * FROM dblist where word like 'sim' or word like 'sim%' or word like '%sim%' or word like '%sim' equal sim is a frist , sim% is second and %sim% is a thrid and then %sim is a last. Currently I can't sort like mysql in sqlite. How to change sql to order the result ?

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  • $_GET encoding problem with cyrillic text

    - by T1000
    I'm trying this code (on my local web server) <?php echo 'the word is / ?????? ? '.$_GET['word']; ?> but I get corrupted result when enter ?word=????? the word is / ?????? ? ???? The document is saved as 'UTF-8 without BOM' and headers are also UTF-8. I have tried urlencode() and urldecode() but the effect was same. When upload it on web server, works fine...

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  • PlaySound linker error in C++

    - by logic-unit
    Hello, I'm getting this error: [Linker error] undefined reference to 'PlaySoundA@12' Id returned 1 exit status From this code: // c++ program to generate a random sequence of numbers then play corresponding audio files #include <windows.h> #include <mmsystem.h> #include <iostream> #pragma comment(lib, "winmm.lib") using namespace std; int main() { int i; i = 0; // set the value of i while (i <= 11) // set the loop to run 11 times { int number; number = rand() % 10 + 1; // generate a random number sequence // cycling through the numbers to find the right wav and play it if (number == 0) { PlaySound("0.wav", NULL, SND_FILENAME); // play the random number } else if (number == 1) { PlaySound("1.wav", NULL, SND_FILENAME); // play the random number } //else ifs repeat to 11... i++; // increment i } return 0; } I've tried absolute and relative paths for the wavs, the file size of them is under 1Mb each too. I've read another thread here on the subject: http://stackoverflow.com/questions/1565439/how-to-playsound-in-c As you may well have guessed this is my first C++ program, so my knowledge is limited with where to go next. I've tried pretty much every page Google has on the subject including MSDN usage page. Any ideas? Thanks in advance...

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  • Highlighting in Solr 1.4 - requireFieldMatch

    - by Mark Redding
    I have an object Title : foo Summary : foo bar Body : this is a published story about a foo and a bar All three are set up as fields with stored=true. The user searches across my system for the word "foo" I would like to highlight foo in all three places. The user searches for the word foo in the title "title:foo" I only want to highlight foo within the title. When I added hl.requireFieldMatch=true and hl.usePhraseHighlighter=true as part of my query over to SOLR I am unable to get the highlighting in all three places when doing a generic non fielded search. Is there a way to get both scenarios to work? I had these two items turned off, but I am adding in some fielded portions of the query that the user does not see which only display Published items for instance. the problem is (foo AND status:published) is causing the word published in the body to highlight when the user only searched for the word "foo".

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  • Rot13 for numbers.

    - by dreeves
    EDIT: Now a Major Motion Blog Post at http://messymatters.com/sealedbids The idea of rot13 is to obscure text, for example to prevent spoilers. It's not meant to be cryptographically secure but to simply make sure that only people who are sure they want to read it will read it. I'd like to do something similar for numbers, for an application involving sealed bids. Roughly I want to send someone my number and trust them to pick their own number, uninfluenced by mine, but then they should be able to reveal mine (purely client-side) when they're ready. They should not require further input from me or any third party. (Added: Note the assumption that the recipient is being trusted not to cheat.) It's not as simple as rot13 because certain numbers, like 1 and 2, will recur often enough that you might remember that, say, 34.2 is really 1. Here's what I'm looking for specifically: A function seal() that maps a real number to a real number (or a string). It should not be deterministic -- seal(7) should not map to the same thing every time. But the corresponding function unseal() should be deterministic -- unseal(seal(x)) should equal x for all x. I don't want seal or unseal to call any webservices or even get the system time (because I don't want to assume synchronized clocks). (Added: It's fine to assume that all bids will be less than some maximum, known to everyone, say a million.) Sanity check: > seal(7) 482.2382 # some random-seeming number or string. > seal(7) 71.9217 # a completely different random-seeming number or string. > unseal(seal(7)) 7 # we always recover the original number by unsealing.

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  • Using repaint() method.

    - by owca
    I'm still struggling to create this game : http://stackoverflow.com/questions/2844190/choosing-design-method-for-ladder-like-word-game .I've got it almost working but there is a problem though. When I'm inserting a word and it's correct, the whole window should reload, and JButtons containing letters should be repainted with different style. But somehow repaint() method for the game panel (in Main method) doesn't affect it at all. What am I doing wrong ? Here's my code: Main: import java.util.Scanner; import javax.swing.*; import java.awt.*; public class Main { public static void main(String[] args){ final JFrame f = new JFrame("Ladder Game"); Scanner sc = new Scanner(System.in); System.out.println("Creating game data..."); System.out.println("Height: "); //setting height of the grid while (!sc.hasNextInt()) { System.out.println("int, please!"); sc.next(); } final int height = sc.nextInt(); /* * I'm creating Grid[]game. Each row of game contains Grid of Element[]line. * Each row of line contains Elements, which are single letters in the game. */ Grid[]game = new Grid[height]; for(int L = 0; L < height; L++){ Grid row = null; int i = L+1; String s; do { System.out.println("Length "+i+", please!"); s = sc.next(); } while (s.length() != i); Element[] line = new Element[s.length()]; Element single = null; String[] temp = null; String[] temp2 = new String[s.length()]; temp = s.split(""); for( int j = temp2.length; j>0; j--){ temp2[j-1] = temp[j]; } for (int k = 0 ; k < temp2.length ; k++) { if( k == 0 ){ single = new Element(temp2[k], 2); } else{ single = new Element(temp2[k], 1); } line[k] = single; } row = new Grid(line); game[L] = row; } //############################################ //THE GAME STARTS HERE //############################################ //create new game panel with box layout JPanel panel = new JPanel(); panel.setLayout(new BoxLayout(panel, BoxLayout.Y_AXIS)); panel.setBackground(Color.ORANGE); panel.setBorder(BorderFactory.createEmptyBorder(10, 10, 10, 10)); //for each row of the game array add panel containing letters Single panel //is drawn with Grid's paint() method and then returned here to be added for(int i = 0; i < game.length; i++){ panel.add(game[i].paint()); } f.setContentPane(panel); f.pack(); f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); f.setVisible(true); boolean end = false; boolean word = false; String text; /* * Game continues until solved() returns true. First check if given word matches the length, * and then the value of any row. If yes - change state of each letter from EMPTY * to OTHER_LETTER. Then repaint the window. */ while( !end ){ while( !word ){ text = JOptionPane.showInputDialog("Input word: "); for(int i = 1; i< game.length; i++){ if(game[i].equalLength(text)){ if(game[i].equalValue(text)){ game[i].changeState(3); f.repaint(); //simple debug - I'm checking if letter, and //state values for each Element are proper for(int k=0; k<=i; k++){ System.out.print(game[k].e[k].letter()); } System.out.println(); for(int k=0; k<=i; k++){ System.out.print(game[k].e[k].getState()); } System.out.println(); //set word to true and ask for another word word = true; } } } } word = false; //check if the game has ended for(int i = 0; i < game.length; i++){ if(game[i].solved()){ end = true; } else { end = false; } } } } } Element: import javax.swing.*; import java.awt.*; public class Element { final int INVISIBLE = 0; final int EMPTY = 1; final int FIRST_LETTER = 2; final int OTHER_LETTER = 3; private int state; private String letter; public Element(){ } //empty block public Element(int state){ this("", 0); } //filled block public Element(String s, int state){ this.state = state; this.letter = s; } public JButton paint(){ JButton button = null; if( state == EMPTY ){ button = new JButton(" "); button.setBackground(Color.WHITE); } else if ( state == FIRST_LETTER ){ button = new JButton(letter); button.setBackground(Color.red); } else { button = new JButton(letter); button.setBackground(Color.yellow); } return button; } public void changeState(int s){ state = s; } public void setLetter(String s){ letter = s; } public String letter(){ return letter; } public int getState(){ return state; } } Grid: import javax.swing.*; import java.awt.*; public class Grid extends JPanel{ public Element[]e; private Grid[]g; public Grid(){} public Grid( Element[]elements ){ e = new Element[elements.length]; for(int i=0; i< e.length; i++){ e[i] = elements[i]; } } public Grid(Grid[]grid){ g = new Grid[grid.length]; for(int i=0; i<g.length; i++){ g[i] = grid[i]; } Dimension d = new Dimension(600, 600); setMinimumSize(d); setPreferredSize(new Dimension(d)); setMaximumSize(d); } //for Each element in line - change state to i public void changeState(int i){ for(int j=0; j< e.length; j++){ e[j].changeState(3); } } //create panel which will be single row of the game. Add elements to the panel. // return JPanel to be added to grid. public JPanel paint(){ JPanel panel = new JPanel(); panel.setLayout(new GridLayout(1, e.length)); panel.setBorder(BorderFactory.createEmptyBorder(2, 2, 2, 2)); for(int j = 0; j < e.length; j++){ panel.add(e[j].paint()); } return panel; } //check if the length of given string is equal to length of row public boolean equalLength(String s){ int len = s.length(); boolean equal = false; for(int j = 0; j < e.length; j++){ if(e.length == len){ equal = true; } } return equal; } //check if the value of given string is equal to values of elements in row public boolean equalValue(String s){ int len = s.length(); boolean equal = false; String[] temp = null; String[] temp2 = new String[len]; temp = s.split(""); for( int j = len; j>0; j--){ temp2[j-1] = temp[j]; } for(int j = 0; j < e.length; j++){ if( e[j].letter().equals(temp2[j]) ){ equal = true; } else { equal = false; } } if(equal){ for(int i = 0; i < e.length; i++){ e[i].changeState(3); } } return equal; } //check if the game has finished public boolean solved(){ boolean solved = false; for(int j = 0; j < e.length; j++){ if(e[j].getState() == 3){ solved = true; } else { solved = false; } } return solved; } }

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  • Creating a fixed formatted cell in UITableview

    - by Wes
    Hi, I want to have a tableview create rows that look like this: value1 item1 container1 value10 item10 container10 value100 item100 container100 value2 item2 container2 What I am trying to show is that the first word (value) will have a set length of 12 and then the second word (item) will have a set length of 10 and then the last word (container) is just tagged on at the end. I am pulling these from a SQLite database and don't want to use multiple lines, but read in a strictly formatted structure like this.

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  • GNU ld removes section

    - by Jonatan
    I'm writing a boot script for an ARM-Cortex M3 based device. If I compile the assembler boot script and the C application code and then combine the object files and transfer them to my device everything works. However, if I use ar to create an archive (libboot.a) and combine that archive with the C application there is a problem: I've put the boot code in a section: .section .boot, "ax" .global _start _start: .word 0x10000800 /* Initial stack pointer (FIXME!) */ .word start .word nmi_handler .word hard_fault_handler ... etc ... I've found that ld strips this from the final binary (the section "boot" is not available). This is quite natural as there is no dependency on it that ld knows about, but it causes the device to not boot correctly. So my question is: what is the best way to force this code to be included?

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  • Select dynamic string has a different value when referenced in Where clause

    - by David
    I dynamically select a string built using another string. So, if string1='David Banner', then MyDynamicString should be 'DBanne' Select ... , Left( left((select top 1 strval from dbo.SPLIT(string1,' ')) //first word ,1) //first character + (select top 1 strval from dbo.SPLIT(string1,' ') //second word where strval not in (select top 1 strval from dbo.SPLIT(string1,' '))) ,6) //1st character of 1st word, followed by up to 5 characters of second word [MyDynamicString] ,... From table1 Join table2 on table1pkey=table2fkey Where MyDynamicString <> table2.someotherfield I know table2.someotherfield is not equal to the dynamic string. However, when I replace MyDynamicString in the Where clause with the full left(left(etc.. function, it works as expected. Can I not reference this string later in the query? Do I have to build it using the left(left(etc.. function each time in the where clause?

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