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  • Consulting a Prolog Source Code from within a VS2008 Solution File

    - by Joshua Green
    I have a Prolog file (Hanoi.pl) containing the code for solving the Hanoi Towers puzzle: hanoi( N ):- move( N, left, middle, right ). move( 0, _, _, _ ):- !. move( N, A, B, C ):- M is N-1, move( M, A, C, B ), inform( A, B ), move( M, C, B, A ). inform( X, Y ):- write( 'move a disk from ' ), write( X ), write( ' to ' ), writeln( Y ). I also have a C++ file written in VS2008 IDE: #include <iostream> #include <string> #include <stdio.h> #include <stdlib.h> using namespace std; #include "SWI-cpp.h" #include "SWI-Prolog.h" predicate_t phanoi; term_t t0; int main(int argc, char** argv) { long n = 5; int rval; if ( !PL_initialise(1, argv) ) PL_halt(1); PL_put_integer( t0, n ); phanoi = PL_predicate( "hanoi", 1, NULL ); rval = PL_call_predicate( NULL, PL_Q_NORMAL, phanoi, t0 ); system( "PAUSE" ); } How can I consult my Prolog source code (Hanoi.pl) from within my C++ code? Not from the Command Prompt - from the code, something like include or consult or compile? It is located in the same folder as my cpp file. Thanks,

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  • Can someone explain me this code ?

    - by VaioIsBorn
    #include <stdio.h> #include <unistd.h> #include <string.h> int good(int addr) { printf("Address of hmm: %p\n", addr); } int hmm() { printf("Win.\n"); execl("/bin/sh", "sh", NULL); } extern char **environ; int main(int argc, char **argv) { int i, limit; for(i = 0; environ[i] != NULL; i++) memset(environ[i], 0x00, strlen(environ[i])); int (*fptr)(int) = good; char buf[32]; if(strlen(argv[1]) <= 40) limit = strlen(argv[1]); for(i = 0; i <= limit; i++) { buf[i] = argv[1][i]; if(i < 36) buf[i] = 0x41; } int (*hmmptr)(int) = hmm; (*fptr)((int)hmmptr); return 0; } I don't really understand the code above, i have it from an online game - i should supply something in the arguments so it would give me shell, but i don't get it how it works so i don't know what to do. So i need someone that would explain it what it does, how it's working and the stuff. Thanks.

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  • How to avoid using the plld.exe utility in VS2008 (for linking C++ and Prolog codes)

    - by Joshua Green
    Here is my code in its entirety: Trying "listing." at the Prolog prompt that pops up when I run the program confirms that my Prolog source code has been loaded (consulted). #include <iostream> #include <fstream> #include <string> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <stdafx.h> using namespace std; #include "Windows.h" #include "ctype.h" #include "SWI-cpp.h" #include "SWI-Prolog.h" #include "SWI-Stream.h" int main(int argc, char** argv) { argc = 4; argv[0] = "libpl.dll"; argv[1] = "-G32m"; argv[2] = "-L32m"; argv[3] = "-T32m"; PL_initialise(argc, argv); if ( !PL_initialise(argc, argv) ) PL_halt(1); PlCall( "consult(swi('plwin.rc'))" ); PlCall( "consult('hello.pl')" ); PL_halt( PL_toplevel() ? 0 : 1 ); } So this is how to load a Prolog source code (hello.pl) at run time into VS2008 without having to use plld at the VS command prompt.

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  • error in finding out the lexems and no of lines of a text file in C

    - by mekasperasky
    #include<stdio.h> #include<ctype.h> #include<string.h> int main() { int i=0,j,k,lines_count[2]={1,1},operand_count[2]={0},operator_count[2]={0},uoperator_count[2]={0},control_count[2]={0,0},cl[13]={0},variable_dec[2]={0,0},l,p[2]={0},ct,variable_used[2]={0,0},constant_count[2],s[2]={0},t[2]={0}; char a,b[100],c[100]; char d[100]={0}; j=30; FILE *fp1[2],*fp2; fp1[0]=fopen("program1.txt","r"); fp1[1]=fopen("program2.txt","r"); //the source file is opened in read only mode which will passed through the lexer fp2=fopen("ccv1ouput.txt","wb"); //now lets remove all the white spaces and store the rest of the words in a file if(fp1[0]==NULL) { perror("failed to open program1.txt"); //return EXIT_FAILURE; } if(fp1[1]==NULL) { perror("failed to open program2.txt"); //return EXIT_FAILURE; } i=0; k=0; ct=0; while(ct!=2) { while(!feof(fp1[ct])) { a=fgetc(fp1[ct]); if(a!=' '&&a!='\n') { if (!isalpha(a) && !isdigit(a)) { switch(a) { case '+':{ i=0; cl[0]=1; operator_count[ct]=operator_count[ct]+1;break;} case '-':{ cl[1]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '*':{ cl[2]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '/':{ cl[3]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '=':{a=fgetc(fp1[ct]); if (a=='='){cl[4]=1; operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1;} else { cl[5]=1; operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; ungetc(1,fp1[ct]); } break;} case '%':{ cl[6]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '<':{ a=fgetc(fp1[ct]); if (a=='=') {cl[7]=1; operator_count[ct]=operator_count[ct]+1;} else { cl[8]=1; operator_count[ct]=operator_count[ct]+1; ungetc(1,fp1[ct]); } break; } case '>':{ ; a=fgetc(fp1[ct]); if (a=='='){cl[9]=1; operator_count[ct]=operator_count[ct]+1;} else { cl[10]=1; operator_count[ct]=operator_count[ct]+1; ungetc(1,fp1[ct]); } break;} case '&':{ cl[11]=1; a=fgetc(fp1[ct]); operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; variable_used[ct]=variable_used[ct]-1; break; } case '|':{ cl[12]=1; a=fgetc(fp1[ct]); operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; variable_used[ct]=variable_used[ct]-1; break; } case '#':{ while(a!='\n') { a=fgetc(fp1[ct]); } } } } else { d[i]=a; i=i+1; k=k+1; } } else { //printf("%s \n",d); if((strcmp(d,"if")==0)){ memset ( d, 0, 100 ); i=0; control_count[ct]=control_count[ct]+1; } else if(strcmp(d,"then")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"else")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"while")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"int")==0){ while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"char")==0){while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"float")==0){while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"printf")==0){while(a!='\n') a=fgetc(fp1[ct]); memset(d,0,100); } else if(strcmp(d,"scanf")==0){while(a!='\n') a=fgetc(fp1[ct]); memset(d,0,100);} else if (isdigit(d[i-1])) { memset ( d, 0, 100 ); i=0; constant_count[ct]=constant_count[ct]+1; operand_count[ct]=operand_count[ct]+1; } else if (isalpha(d[i-1]) && strcmp(d,"int")!=0 && strcmp(d,"char")!=0 && strcmp(d,"float")!=0 && (strcmp(d,"if")!=0) && strcmp(d,"then")!=0 && strcmp(d,"else")!=0 && strcmp(d,"while")!=0 && strcmp(d,"printf")!=0 && strcmp(d,"scanf")!=0) { memset ( d, 0, 100 ); i=0; operand_count[ct]=operand_count[ct]+1; } else if(a=='\n') { lines_count[ct]=lines_count[ct]+1; memset ( d, 0, 100 ); } } } fclose(fp1[ct]); operand_count[ct]=operand_count[ct]-5; variable_used[0]=operand_count[0]-constant_count[0]; variable_used[1]=operand_count[1]-constant_count[1]; for(j=0;j<12;j++) uoperator_count[ct]=uoperator_count[ct]+cl[j]; fprintf(fp2,"\n statistics of program %d",ct+1); fprintf(fp2,"\n the no of lines ---> %d",lines_count[ct]); fprintf(fp2,"\n the no of operands --->%d",operand_count[ct]); fprintf(fp2,"\n the no of operator --->%d",operator_count[ct]); fprintf(fp2,"\n the no of control statments --->%d",control_count[ct]); fprintf(fp2,"\n the no of unique operators --->%d",uoperator_count[ct]); fprintf(fp2,"\n the no of variables declared--->%d",variable_dec[ct]); fprintf(fp2,"\n the no of variables used--->%d",variable_used[ct]); fprintf(fp2,"\n ---------------------------------"); fprintf(fp2,"\n \t \t \t"); ct=ct+1; } t[0]=lines_count[0]+control_count[0]+uoperator_count[0]; t[1]=lines_count[1]+control_count[1]+uoperator_count[1]; s[0]=operator_count[0]+operand_count[0]+variable_dec[0]+variable_used[0]; s[1]=operator_count[1]+operand_count[1]+variable_dec[1]+variable_used[1]; fprintf(fp2,"\n the time complexity of program 1 is %d",t[0]); fprintf(fp2,"\n the time complexity of program 2 is %d",t[1]); fprintf(fp2,"\n the space complexity of program 1 is %d",s[0]); fprintf(fp2,"\n the space complexity of program 2 is %d",s[1]); if((t[0]>t[1]) && (s[0] >s[1])) fprintf(fp2,"\n the efficiency of program 2 is greater than program 1"); else if(t[0]<t[1] && s[0] < s[1]) fprintf(fp2,"\n the efficiency of program 1 is greater than program 2 " ); else if (t[0]+s[0] > t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 1 is greater than program 2"); else if (t[0]+s[0] < t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 2 is greater than program 1"); else if (t[0]+s[0] == t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 1 is equal to that of program 2"); fclose(fp2); return 0; } this code basically compares two c codes and finds out the no. of variables declared , used , no. of control statements , no. of lines and no. of unique operators , and operands , so as to find out the time complexity and space complexity of of the two programs given in the text file program1.txt and program2.txt ... Lets say program1.txt is this #include<stdio.h> #include<math.h> int main () { FILE *fp; fp=fopen("output.txt","w"); long double t,y=0,x=0,e=5,f=1,w=1; for (t=0;t<10;t=t+0.01) { //if (isnan(y) || isinf(y)) //break; fprintf(fp,"%ld\t%ld\n",y,x); y = y + ((e*(1 - (x*x))*y) - x + f*cos(w*0.1))*0.1; x = x + y*0.1; } fclose(fp); return (0); } i havent indented it as its just a text file . But my output is totally faulty . Its not able to find the any of the ouput that i need . Where is the bug in this ? I am not able to figure out as the algorithm looks fine .

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  • How can I declare and initialize an array of pointers to a structure in C?

    - by worlds-apart89
    I have a small assignment in C. I am trying to create an array of pointers to a structure. My question is how can I initialize each pointer to NULL? Also, after I allocate memory for a member of the array, I can not assign values to the structure to which the array element points. #include <stdio.h> #include <stdlib.h> typedef struct list_node list_node_t; struct list_node { char *key; int value; list_node_t *next; }; int main() { list_node_t *ptr = (list_node_t*) malloc(sizeof(list_node_t)); ptr->key = "Hello There"; ptr->value = 1; ptr->next = NULL; // Above works fine // Below is erroneous list_node_t **array[10] = {NULL}; *array[0] = (list_node_t*) malloc(sizeof(list_node_t)); array[0]->key = "Hello world!"; //request for member ‘key’ in something not a structure or union array[0]->value = 22; //request for member ‘value’ in something not a structure or union array[0]->next = NULL; //request for member ‘next’ in something not a structure or union // Do something with the data at hand // Deallocate memory using function free return 0; }

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  • g++ Linking Error on Mac while compiling FFMPEG

    - by Saptarshi Biswas
    g++ on Snow Leopard is throwing linking errors on the following piece of code test.cpp #include <iostream> using namespace std; #include <libavcodec/avcodec.h> // required headers #include <libavformat/avformat.h> int main(int argc, char**argv) { av_register_all(); // offending library call return 0; } When I try to compile this using the following command g++ test.cpp -I/usr/local/include -L/usr/local/lib \ -lavcodec -lavformat -lavutil -lz -lm -o test I get the error Undefined symbols: "av_register_all()", referenced from: _main in ccUD1ueX.o ld: symbol(s) not found collect2: ld returned 1 exit status Interestingly, if I have an equivalent c code, test.c #include <stdio.h> #include <libavcodec/avcodec.h> #include <libavformat/avformat.h> int main(int argc, char**argv) { av_register_all(); return 0; } gcc compiles it just fine gcc test.c -I/usr/local/include -L/usr/local/lib \ -lavcodec -lavformat -lavutil -lz -lm -o test I am using Mac OS X 10.6.5 $ g++ --version i686-apple-darwin10-g++-4.2.1 (GCC) 4.2.1 (Apple Inc. build 5664) $ gcc --version i686-apple-darwin10-gcc-4.2.1 (GCC) 4.2.1 (Apple Inc. build 5664) FFMPEG's libavcodec, libavformat etc. are C libraries and I have built them on my machine like thus: ./configure --enable-gpl --enable-pthreads --enable-shared \ --disable-doc --enable-libx264 make && sudo make install As one would expect, libavformat indeed contains the symbol av_register_all $ nm /usr/local/lib/libavformat.a | grep av_register_all 0000000000000000 T _av_register_all 00000000000089b0 S _av_register_all.eh I am inclined to believe g++ and gcc have different views of the libraries on my machine. g++ is not able to pick up the right libraries. Any clue?

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  • Help me with the simplest program for "Trusted" application

    - by idazuwaika
    Hi, I hope anyone from the large community here can help me write the simplest "Trusted" program that I can expand from. I'm using Ubuntu Linux 9.04, with TPM emulator 0.60 from Mario Strasser (http://tpm-emulator.berlios.de/). I have installed the emulator and Trousers, and can successfully run programs from tpm-tools after running tpmd and tcsd daemons. I hope to start developing my application, but I have problems compiling the code below. #include <trousers/tss.h> #include <trousers/trousers.h> #include <stdio.h> TSS_HCONTEXT hContext; int main() { Tspi_Context_Create(&hContext); Tspi_Context_Close(hContext); return 0; } After trying to compile with g++ tpm.cpp -o tpmexe I receive errors undefined reference to 'Tspi_Context_Create' undefined reference to 'Tspi_Context_Close' What do I have to #include to successfully compile this? Is there anything that I miss? I'm familiar with C, but not exactly so with Linux/Unix programming environment. ps: I am a part time student in Master in Information Security programme. My involvement with programming has been largely for academic purposes.

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  • How do calculators work with precision?

    - by zoul
    Hello! I wonder how calculators work with precision. For example the value of sin(M_PI) is not exactly zero when computed in double precision: #include <math.h> #include <stdio.h> int main() { double x = sin(M_PI); printf("%.20f\n", x); // 0.00000000000000012246 return 0; } Now I would certainly want to print zero when user enters sin(p). I can easily round somewhere on 1e–15 to make this particular case work, but that’s a hack, not a solution. When I start to round like this and the user enters something like 1e–20, they get a zero back (because of the rounding). The same thing happens when the user enters 1/10 and hits the = key repeatedly — when he reaches the rounding treshold, he gets zero. And yet some calculators return plain zero for sin(p) and at the same time they can work with expressions such as (1e–20)/10 comfortably. Where’s the trick?

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  • Doubts in executable and relocatable object file

    - by bala1486
    Hello, I have written a simple Hello World program. #include <stdio.h> int main() { printf("Hello World"); return 0; } I wanted to understand how the relocatable object file and executable file look like. The object file corresponding to the main function is 0000000000000000 <main>: 0: 55 push %rbp 1: 48 89 e5 mov %rsp,%rbp 4: bf 00 00 00 00 mov $0x0,%edi 9: b8 00 00 00 00 mov $0x0,%eax e: e8 00 00 00 00 callq 13 <main+0x13> 13: b8 00 00 00 00 mov $0x0,%eax 18: c9 leaveq 19: c3 retq Here the function call for printf is callq 13. One thing i don't understand is why is it 13. That means call the function at adresss 13, right??. 13 has the next instruction, right?? Please explain me what does this mean?? The executable code corresponding to main is 00000000004004cc <main>: 4004cc: 55 push %rbp 4004cd: 48 89 e5 mov %rsp,%rbp 4004d0: bf dc 05 40 00 mov $0x4005dc,%edi 4004d5: b8 00 00 00 00 mov $0x0,%eax 4004da: e8 e1 fe ff ff callq 4003c0 <printf@plt> 4004df: b8 00 00 00 00 mov $0x0,%eax 4004e4: c9 leaveq 4004e5: c3 retq Here it is callq 4003c0. But the binary instruction is e8 e1 fe ff ff. There is nothing that corresponds to 4003c0. What is that i am getting wrong? Thanks. Bala

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  • Xlib.h and Xutil.h not found in Eclipse, how can I fix this?

    - by eclipseNoob
    Hi, I'm a newbie to Eclipse IDE for C/C++ development. I just installed MingW and set it up as my system's environment variable. I am trying to make an application that uses the X library but eclipse cant seem to find it. Eclipse works with any other simple standard library functions but it cant find the X library. Please Help! Here's a sample code snippet that's failing: #include <stdio.h> #include <X11/Xlib.h> // Can't find this #include <X11/Xutil.h> // Or this... int main() { printf("Hello"); return 0; } Do I have to download the X library from somewhere? If so then from where and where do I paste it to? Please tell me what to do in order for me to start coding using the Xlib in Eclipse. If you find any useful links, please dont hesitate to post. Thanks.

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  • Incompatible types when assigning to type 'struct compartido'

    - by user1660559
    I have one problem with this code. I should create one structure and share it across 5 new process created from the father: #include <stdio.h> #include <stdlib.h> #include <sys/wait.h> #include <unistd.h> #include <sys/types.h> #include <sys/ipc.h> #include <sys/shm.h> #include <sys/sem.h> #include <time.h> struct compartido { int pid1, pid2, pid3, pid4, pid5; int propietario; int contador; int pidpadre; }; struct compartido var; int main(int argc, char *argv[]) { key_t llave1,llavesem; int idmem,idsem; llave1=ftok("/tmp",'a'); idmem=shmget(llave1,sizeof(int),IPC_CREAT|0600); if (idmem==-1) { perror ("shmget"); return 1; } var=shmat(idmem,0,0); /*This line is giving the error*/ /*rest of the code*/ } The exact error is giving is: error: incompatible types when assigning to type 'struct compartido' from type 'void *' I need to put this structure in the shared variable to be able to see and modify all those data from the 6 process (5 children and the father). What I'm doing bad? Thanks in advance and best regards,

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  • Equivalent to window.setTimeout() for C++

    - by bobobobo
    In javascript there's this sweet, sweet function window.setTimeout( func, 1000 ) ; which will asynchronously invoke func after 1000 ms. I want to do something similar in C++ (without multithreading), so I put together a sample loop like: #include <stdio.h> struct Callback { // The _time_ this function will be executed. double execTime ; // The function to execute after execTime has passed void* func ; } ; // Sample function to execute void go() { puts( "GO" ) ; } // Global program-wide sense of time double time ; int main() { // start the timer time = 0 ; // Make a sample callback Callback c1 ; c1.execTime = 10000 ; c1.func = go ; while( 1 ) { // its time to execute it if( time c1.execTime ) { c1.func ; // !! doesn't work! } time++; } } How can I make something like this work?

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  • Kernighan & Ritchie word count example program in a functional language

    - by Frank
    I have been reading a little bit about functional programming on the web lately and I think I got a basic idea about the concepts behind it. I'm curious how everyday programming problems which involve some kind of state are solved in a pure functional programing language. For example: how would the word count program from the book 'The C programming Language' be implemented in a pure functional language? Any contributions are welcome as long as the solution is in a pure functional style. Here's the word count C code from the book: #include <stdio.h> #define IN 1 /* inside a word */ #define OUT 0 /* outside a word */ /* count lines, words, and characters in input */ main() { int c, nl, nw, nc, state; state = OUT; nl = nw = nc = 0; while ((c = getchar()) != EOF) { ++nc; if (c == '\n') ++nl; if (c == ' ' || c == '\n' || c = '\t') state = OUT; else if (state == OUT) { state = IN; ++nw; } } printf("%d %d %d\n", nl, nw, nc); }

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  • pthread_exit and/or pthread_join causing Abort and SegFaults.

    - by MJewkes
    The following code is a simple thread game, that switches between threads causing the timer to decrease. It works fine for 3 threads, causes and Abort(core dumped) for 4 threads, and causes a seg fault for 5 or more threads. Anyone have any idea why this might be happening? #include <stdio.h> #include <stdlib.h> #include <pthread.h> #include <errno.h> #include <assert.h> int volatile num_of_threads; int volatile time_per_round; int volatile time_left; int volatile turn_id; int volatile thread_running; int volatile can_check; void * player (void * id_in){ int id= (int)id_in; while(1){ if(can_check){ if (time_left<=0){ break; } can_check=0; if(thread_running){ if(turn_id==id-1){ turn_id=random()%num_of_threads; time_left--; } } can_check=1; } } pthread_exit(NULL); } int main(int argc, char *args[]){ int i; int buffer; pthread_t * threads =(pthread_t *)malloc(num_of_threads*sizeof(pthread_t)); thread_running=0; num_of_threads=atoi(args[1]); can_check=0; time_per_round = atoi(args[2]); time_left=time_per_round; srandom(time(NULL)); //Create Threads for (i=0;i<num_of_threads;i++){ do{ buffer=pthread_create(&threads[i],NULL,player,(void *)(i+1)); }while(buffer == EAGAIN); } can_check=1; time_left=time_per_round; turn_id=random()%num_of_threads; thread_running=1; for (i=0;i<num_of_threads;i++){ assert(!pthread_join(threads[i], NULL)); } return 0; }

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  • Newb Question: scanf() in C

    - by riemannliness
    So I started learning C today, and as an exercise i was told to write a program that asks the user for numbers until they type a 0, then adds the even ones and the odd ones together. Here is is (don't laugh at my bad style): #include <stdio.h>; int main() { int esum = 0, osum = 0; int n, mod; puts("Please enter some numbers, 0 to terminate:"); scanf("%d", &n); while (n != 0) { mod = n % 2; switch(mod) { case 0: esum += n; break; case 1: osum += n; } scanf("%d", &n); } printf("The sum of evens:%d,\t The sum of odds:%d", esum, osum); return 0; } My question concerns the mechanics of the scanf() function. It seems that when you enter several numbers at once separated by spaces (eg. 1 22 34 2 8), the scanf() function somehow remembers each distinct numbers in the line, and steps through the while loop for each one respectively. Why/how does this happen? Example interaction within command prompt: - Please enter some numbers, 0 to terminate: 42 8 77 23 11 (enter) 0 (enter) - The sum of evens:50, The sum of odds:111 I'm running the program through the command prompt, it's compiled for win32 platforms with visual studio.

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  • dynamic lib can't find static lib

    - by renyufei
    env: gcc version 4.4.1 (Ubuntu 4.4.1-4ubuntu9) app: Bin(main) calls dynamic lib(testb.so), and testb.so contains a static lib(libtesta.a). file list: main.c test.h a.c b.c then compile as: gcc -o testa.o -c a.c ar -r libtesta.a testa.o gcc -shared -fPIC -o testb.so b.c gcc -o main main.c -L. -ltesta -ldl then compile success, but runs an error: ./main: symbol lookup error: ./testb.so: undefined symbol: print code as follows: test.h #include <stdio.h> #include <stdlib.h> #include <errno.h> #include <string.h> #include <dlfcn.h> int printa(const char *msg); int printb(const char *msg); a.c #include "test.h" int printa(const char *msg) { printf("\tin printa\n"); printf("\t%s\n", msg); } b.c #include "test.h" int printb(const char *msg) { printf("in printb\n"); printa("called by printb\n"); printf("%s\n", msg); } main.c #include "test.h" int main(int argc, char **argv) { void *handle; int (*dfn)(const char *); printf("before dlopen\n"); handle = dlopen("./testb.so", RTLD_LOCAL | RTLD_LAZY); printf("after dlopen\n"); if (handle == NULL) { printf("dlopen fail: [%d][%s][%s]\n", \ errno, strerror(errno), dlerror()); exit(EXIT_FAILURE); } printf("before dlsym\n"); dfn = dlsym(handle, "printb"); printf("after dlsym\n"); if (dfn == NULL) { printf("dlsym fail: [%d][%s][%s]\n", \ errno, strerror(errno), dlerror()); exit(EXIT_FAILURE); } printf("before dfn\n"); dfn("printb func\n"); printf("after dfn\n"); exit(EXIT_SUCCESS); }

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  • C++ Check Substring of a String

    - by user69514
    I'm trying to check whether or not the second argument in my program is a substring of the first argument. The problem is that it only work if the substring starts with the same letter of the string. .i.e Michigan - Mich (this works) Michigan - Mi (this works) Michigan - igan (this doesn't work) #include <stdio.h> #include <string.h> #include <string> using namespace std; bool my_strstr( string str, string sub ) { bool flag = true; int startPosition = -1; char subStart = str.at(0); char strStart; //find starting position for(int i=0; i<str.length(); i++){ if(str.at(i) == subStart){ startPosition = i; break; } } for(int i=0; i<sub.size(); i++){ if(sub.at(i) != str.at(startPosition)){ flag = false; break; } startPosition++; } return flag; } int main(int argc, char **argv){ if (argc != 3) { printf ("Usage: check <string one> <string two>\n"); } string str1 = argv[1]; string str2 = argv[2]; bool result = my_strstr(str1, str2); if(result == 1){ printf("%s is a substring of %s\n", argv[2], argv[1]); } else{ printf("%s is not a substring of %s\n", argv[2], argv[1]); } return 0; }

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  • C89, Mixing Variable Declarations and Code

    - by rutski
    I'm very curious to know why exactly C89 compilers will dump on you when you try to mix variable declarations and code, like this for example: rutski@imac:~$ cat test.c #include <stdio.h> int main(void) { printf("Hello World!\n"); int x = 7; printf("%d!\n", x); return 0; } rutski@imac:~$ gcc -std=c89 -pedantic test.c test.c: In function ‘main’: test.c:7: warning: ISO C90 forbids mixed declarations and code rutski@imac:~$ Yes, you can avoid this sort of thing by staying away from -pedantic. But then your code is no longer standards compliant. And as anybody capable of answering this post probably already knows, this is not just a theoretical concern. Platforms like Microsoft's C compiler enforce this quick in the standard under any and all circumstances. Given how ancient C is, I would imagine that this feature is due to some historical issue dating back to the extraordinary hardware limitations of the 70's, but I don't know the details. Or am I totally wrong there?

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  • Hi I am facing a fragmentation error while executing this code? Can someone explain why?

    - by aks
    #include<stdio.h> struct table { char *ipAddress; char *domainName; struct table *next; }; struct table *head = NULL; void add_rec(); void show_rec(); int main() { add_rec(); show_rec(); return 0; } void add_rec() { struct table * temp = head; struct table * temp1 = (struct table *)malloc(sizeof(struct table)); if(!temp1) printf("\n Unable to allocate memory \n"); printf("Enter the ip address you want \n"); scanf("%s",temp1->ipAddress); printf("\nEnter the domain name you want \n"); scanf("%s",temp1->domainName); if(!temp) { head = temp; } else { while(temp->next!=NULL) temp = temp->next; temp->next = temp1; } } void show_rec() { struct table * temp = head; if(!temp) printf("\n No entry exists \n"); while(temp!=NULL) { printf("ipAddress = %s\t domainName = %s\n",temp->ipAddress,temp->domainName); temp = temp->next; } } When i execute this code and enters the IP address for the first node, i am facing fragmentation error. The code crashed. Can someone enlighten?

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  • select failing with C program but not shell

    - by Gary
    So I have a parent and child process, and the parent can read output from the child and send to the input of the child. So far, everything has been working fine with shell scripts, testing commands which input and output data. I just tested with a simple C program and couldn't get it to work. Here's the C program: #include <stdio.h> int main( void ) { char stuff[80]; printf("Enter some stuff:\n"); scanf("%s", stuff); return 0; } The problem with with the C program is that my select fails to read from the child fd and hence the program cannot finish. Here's the bit that does the select.. //wait till child is ready fd_set set; struct timeval timeout; FD_ZERO( &set ); // initialize fd set FD_SET( PARENT_READ, &set ); // add child in to set timeout.tv_sec = 3; timeout.tv_usec = 0; int r = select(FD_SETSIZE, &set, NULL, NULL, &timeout); if( r < 1 ) { // we didn't get any input exit(1); } Does anyone have any idea why this would happen with the C program and not a shell one? Thanks!

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  • program won't find math.h anymore

    - by 130490868091234
    After a long time, I downloaded a program I co-developed and tried to recompile it on my Ubuntu Linux 12.04, but it seems it does not find math.h anymore. This may be because something has changed recently in gcc, but I can't figure out if it's something wrong in src/Makefile.am or a missing dependency: Download from http://www.ub.edu/softevol/variscan/: tar xzf variscan-2.0.2.tar.gz cd variscan-2.0.2/ make distclean sh ./autogen.sh make I get: [...] gcc -DNDEBUG -O3 -W -Wall -ansi -pedantic -lm -o variscan variscan.o statistics.o common.o linefile.o memalloc.o dlist.o errabort.o dystring.o intExp.o kxTok.o pop.o window.o free.o output.o readphylip.o readaxt.o readmga.o readmaf.o readhapmap.o readxmfa.o readmav.o ran1.o swcolumn.o swnet.o swpoly.o swref.o statistics.o: In function `calculate_Fu_and_Li_D': statistics.c:(.text+0x497): undefined reference to `sqrt' statistics.o: In function `calculate_Fu_and_Li_F': statistics.c:(.text+0x569): undefined reference to `sqrt' statistics.o: In function `calculate_Fu_and_Li_D_star': statistics.c:(.text+0x63b): undefined reference to `sqrt' statistics.o: In function `calculate_Fu_and_Li_F_star': statistics.c:(.text+0x75c): undefined reference to `sqrt' statistics.o: In function `calculate_Tajima_D': statistics.c:(.text+0x85d): undefined reference to `sqrt' statistics.o:statistics.c:(.text+0xcb1): more undefined references to `sqrt' follow statistics.o: In function `calcRunMode21Stats': statistics.c:(.text+0xe02): undefined reference to `log' statistics.o: In function `correctedDivergence': statistics.c:(.text+0xe5a): undefined reference to `log' statistics.o: In function `calcRunMode22Stats': statistics.c:(.text+0x104a): undefined reference to `sqrt' statistics.o: In function `calculate_Fu_fs': statistics.c:(.text+0x11a8): undefined reference to `fabsl' statistics.c:(.text+0x11ca): undefined reference to `powl' statistics.c:(.text+0x11f2): undefined reference to `logl' statistics.o: In function `calculateStatistics': statistics.c:(.text+0x13f2): undefined reference to `log' collect2: ld returned 1 exit status make[1]: *** [variscan] Error 1 make[1]: Leaving directory `/home/avilella/variscan/latest/variscan-2.0.2/src' make: *** [all-recursive] Error 1 The libraries are there because this simple example works perfectly well: $ gcc test.c -o test -lm $ cat test.c #include <stdio.h> #include <math.h> int main(void) { double x = 0.5; double result = sqrt(x); printf("The hyperbolic cosine of %lf is %lf\n", x, result); return 0; } Any ideas?

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  • Error when linking C executable to OpenCV

    - by Ghilas BELHADJ
    I'm compiling OpenCV under Ubuntu 13.10 using cMake. i've already compiled c++ programs and they works well. now i'm trying to compile a C file using this cMakeLists.txt cmake_minimum_required (VERSION 2.8) project (hello) find_package (OpenCV REQUIRED) add_executable (hello src/test.c) target_link_libraries (hello ${OpenCV_LIBS}) here is the test.c file: #include <stdio.h> #include <stdlib.h> #include <opencv/highgui.h> int main (int argc, char* argv[]) { IplImage* img = NULL; const char* window_title = "Hello, OpenCV!"; if (argc < 2) { fprintf (stderr, "usage: %s IMAGE\n", argv[0]); return EXIT_FAILURE; } img = cvLoadImage(argv[1], CV_LOAD_IMAGE_UNCHANGED); if (img == NULL) { fprintf (stderr, "couldn't open image file: %s\n", argv[1]); return EXIT_FAILURE; } cvNamedWindow (window_title, CV_WINDOW_AUTOSIZE); cvShowImage (window_title, img); cvWaitKey(0); cvDestroyAllWindows(); cvReleaseImage(&img); return EXIT_SUCCESS; } it returns me this error whene running cmake . then make to the project: Linking C executable hello /usr/bin/ld: CMakeFiles/hello.dir/src/test.c.o: undefined reference to symbol «lrint@@GLIBC_2.1» /lib/i386-linux-gnu/libm.so.6: error adding symbols: DSO missing from command line collect2: error: ld returned 1 exit status make[2]: *** [hello] Erreur 1 make[1]: *** [CMakeFiles/hello.dir/all] Erreur 2 make: *** [all] Erreur 2

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  • Hide struct definition in static library.

    - by BobMcLaury
    Hi, I need to provide a C static library to the client and need to be able to make a struct definition unavailable. On top of that I need to be able to execute code before the main at library initialization using a global variable. Here's my code: private.h #ifndef PRIVATE_H #define PRIVATE_H typedef struct TEST test; #endif private.c (this should end up in a static library) #include "private.h" #include <stdio.h> struct TEST { TEST() { printf("Execute before main and have to be unavailable to the user.\n"); } int a; // Can be modified by the user int b; // Can be modified by the user int c; // Can be modified by the user } TEST; main.c test t; int main( void ) { t.a = 0; t.b = 0; t.c = 0; return 0; } Obviously this code doesn't work... but show what I need to do... Anybody knows how to make this work? I google quite a bit but can't find an answer, any help would be greatly appreciated. TIA!

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  • how to fix my error saying expected expression before 'else'

    - by user292489
    this program intended to read a .txt, a set of numbers, file and wwrite to another two .txt files called even amd odd as follows: #include <stdio.h> #include <stdlib.h> int main(int argc, char *argv[]) { int i=0,even,odd; int number[i]; // check to make sure that all the file names are entered if (argc != 3) { printf("Usage: executable in_file output_file\n"); exit(0); } FILE *dog = fopen(argv[1], "r"); FILE *feven= fopen(argv[2], "w"); FILE *fodd= fopen (argv[3], "w"); // check whether the file has been opened successfully if (dog == NULL) { printf("File %s cannot open!\n", argv[1]); exit(0); } //odd = fopen(argv[2], "w"); { if (i%2!=1) i++;} fprintf(feven, "%d", even); fscanf(dog, "%d", &number[i]); else { i%2==1; i++;} fprintf(fodd, "%d", odd); fscanf(dog, "%d", &number[i]); fclose(feven); fclose(fodd);

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  • error: invalid type argument of '->' (have 'struct node')

    - by Roshan S.A
    Why cant i access the pointer "Cells" like an array ? i have allocated the appropriate memory why wont it act like an array here? it works like an array for a pointer of basic data types. #include<stdio.h> #include<stdlib.h> #include<ctype.h> #define MAX 10 struct node { int e; struct node *next; }; typedef struct node *List; typedef struct node *Position; struct Hashtable { int Tablesize; List Cells; }; typedef struct Hashtable *HashT; HashT Initialize(int SIZE,HashT H) { int i; H=(HashT)malloc(sizeof(struct Hashtable)); if(H!=NULL) { H->Tablesize=SIZE; printf("\n\t%d",H->Tablesize); H->Cells=(List)malloc(sizeof(struct node)* H->Tablesize); should it not act like an array from here on? if(H->Cells!=NULL) { for(i=0;i<H->Tablesize;i++) the following lines are the ones that throw the error { H->Cells[i]->next=NULL; H->Cells[i]->e=i; printf("\n %d",H->Cells[i]->e); } } } else printf("\nError!Out of Space"); } int main() { HashT H; H=Initialize(10,H); return 0; } The error I get is as in the title-error: invalid type argument of '->' (have 'struct node').

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