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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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• 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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• 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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• Accessing memory buffer after fread()

  - by xiongtx

  I'm confused as to how fread() is used. Below is an example from cplusplus.com /* fread example: read a complete file */ #include <stdio.h> #include <stdlib.h> int main () { FILE * pFile; long lSize; char * buffer; size_t result; pFile = fopen ( "myfile.bin" , "rb" ); if (pFile==NULL) {fputs ("File error",stderr); exit (1);} // obtain file size: fseek (pFile , 0 , SEEK_END); lSize = ftell (pFile); rewind (pFile); // allocate memory to contain the whole file: buffer = (char*) malloc (sizeof(char)*lSize); if (buffer == NULL) {fputs ("Memory error",stderr); exit (2);} // copy the file into the buffer: result = fread (buffer,1,lSize,pFile); if (result != lSize) {fputs ("Reading error",stderr); exit (3);} /* the whole file is now loaded in the memory buffer. */ // terminate fclose (pFile); free (buffer); return 0; } Let's say that I don't use fclose() just yet. Can I now just treat buffer as an array and access elements like buffer[i]? Or do I have to do something else?

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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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• 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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• Segmentation fault

  - by darkie15

  #include<stdio.h> #include<zlib.h> #include<unistd.h> #include<string.h> int main(int argc, char *argv[]) { char *path=NULL; size_t size; int index ; printf("\nArgument count is = %d", argc); printf ("\nThe 0th argument to the file is %s", argv[0]); path = getcwd(path, size); printf("\nThe current working directory is = %s", path); if (argc <= 1) { printf("\nUsage: ./output filename1 filename2 ..."); } else if (argc > 1) { for (index = 1; index <= argc;index++) { printf("\n File name entered is = %s", argv[index]); strcat(path,argv[index]); printf("\n The complete path of the file name is = %s", path); } } return 0; } In the above code, here is the output that I get while running the code: $ ./output test.txt Argument count is = 2 The 0th argument to the file is ./output The current working directory is = /home/welcomeuser File name entered is = test.txt The complete path of the file name is = /home/welcomeusertest.txt Segmentation fault (core dumped) Can anyone please me understand why I am getting a core dumped error? Regards, darkie

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• Cant print contents of a custom file

  - by ZaZu

  Hello, Im trying to scan contents from a random file into an array in a structure. Then I want to print those contents on screen. (NOTE: The following code is from a bigger program, this is just a sample, but all structures and arrays used are needed as declared ) The contents of the file being tested are simply: 5 4 3 2 5 3 4 2 #include<stdio.h> #define first 500 #define sec 500 struct trial{ int f; int r; float what[first][sec]; }; int trialtest(trial *test); int trialdisplay(trial *test); main(){ trial test; trialtest(&test); trialdisplay(&test); } int trialtest(trial *test){ int z,x,i; FILE *inputf; inputf=fopen("randomfile.txt","r"); for(i=0;i<5;i++){ fscanf(inputf,"%f",&(*test).what[z][x]); } fclose(inputf); return 0; } int trialdisplay(trial *test){ int i,z,x; printf("printing\n\n\n"); for (i=0;i<10;i++){ printf("%f",(*test).what[z][x]); } return 0; } The problem is, I get this error whenever I run the code .. I cant really understand whats going on : Any suggestions ? Thanks alot !

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• function pointers callbacks C

  - by robUK

  Hello, I have started to review callbacks. I found this link: http://stackoverflow.com/questions/142789/what-is-a-callback-in-c-and-how-are-they-implemented which has a good example of callback which is very similar to what we use at work. However, I have tried to get it to work, but I have many errors. #include <stdio.h> /* Is the actual function pointer? */ typedef void (*event_cb_t)(const struct event *evt, void *user_data); struct event_cb { event_cb_t cb; void *data; }; int event_cb_register(event_ct_t cb, void *user_data); static void my_event_cb(const struct event *evt, void *data) { /* do some stuff */ } int main(void) { event_cb_register(my_event_cb, &my_custom_data); struct event_cb *callback; callback->cb(event, callback->data); return 0; } I know that callback use function pointers to store an address of a function. But there is a few things that I find I don't understand. That is what is meet by "registering the callback" and "event dispatcher"? Many thanks for any advice,

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• C Programing - Return libcurl http response to string to the calling function

  - by empty set

  I have a homework and i need somehow to compare two http responses. I am writing it on C (dumb decision) and i use libcurl to make things easier. I am calling the function that uses libcurl to http request and response from another function and i want to return the http response to it. Anyway, the code below doesn't work, any ideas? #include <stdio.h> #include <curl/curl.h> #include <string.h> size_t write_data(void *ptr, size_t size, size_t nmemb, void *stream) { size_t written; written = fwrite(ptr, size, nmemb, stream); return written; } char *handle_url(void) { CURL *curl; char *fp; CURLcode res; char *url = "http://www.yahoo.com"; curl = curl_easy_init(); if (curl) { curl_easy_setopt(curl, CURLOPT_URL, url); curl_easy_setopt(curl, CURLOPT_WRITEFUNCTION, write_data); curl_easy_setopt(curl, CURLOPT_WRITEDATA, fp); res = curl_easy_perform(curl); if(res != CURLE_OK) fprintf(stderr, "curl_easy_perform() failed: %s\n", curl_easy_strerror(res)); curl_easy_cleanup(curl); //printf("\n%s", fp); } return fp; } This solution C libcurl get output into a string works, but not in my case because i want to return this string to the calling function.

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• What limits scaling in this simple OpenMP program?

  - by Douglas B. Staple

  I'm trying to understand limits to parallelization on a 48-core system (4xAMD Opteron 6348, 2.8 Ghz, 12 cores per CPU). I wrote this tiny OpenMP code to test the speedup in what I thought would be the best possible situation (the task is embarrassingly parallel): // Compile with: gcc scaling.c -std=c99 -fopenmp -O3 #include <stdio.h> #include <stdint.h> int main(){ const uint64_t umin=1; const uint64_t umax=10000000000LL; double sum=0.; #pragma omp parallel for reduction(+:sum) for(uint64_t u=umin; u<umax; u++) sum+=1./u/u; printf("%e\n", sum); } I was surprised to find that the scaling is highly nonlinear. It takes about 2.9s for the code to run with 48 threads, 3.1s with 36 threads, 3.7s with 24 threads, 4.9s with 12 threads, and 57s for the code to run with 1 thread. Unfortunately I have to say that there is one process running on the computer using 100% of one core, so that might be affecting it. It's not my process, so I can't end it to test the difference, but somehow I doubt that's making the difference between a 19~20x speedup and the ideal 48x speedup. To make sure it wasn't an OpenMP issue, I ran two copies of the program at the same time with 24 threads each (one with umin=1, umax=5000000000, and the other with umin=5000000000, umax=10000000000). In that case both copies of the program finish after 2.9s, so it's exactly the same as running 48 threads with a single instance of the program. What's preventing linear scaling with this simple program?

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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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• tail call generated by clang 1.1 and 1.0 (llvm 2.7 and 2.6)

  - by ony

  After compilation next snippet of code with clang -O2 (or with online demo): #include <stdio.h> #include <stdlib.h> int flop(int x); int flip(int x) { if (x == 0) return 1; return (x+1)*flop(x-1); } int flop(int x) { if (x == 0) return 1; return (x+0)*flip(x-1); } int main(int argc, char **argv) { printf("%d\n", flip(atoi(argv[1]))); } I'm getting next snippet of llvm assembly in flip: bb1.i: ; preds = %bb1 %4 = add nsw i32 %x, -2 ; <i32> [#uses=1] %5 = tail call i32 @flip(i32 %4) nounwind ; <i32> [#uses=1] %6 = mul nsw i32 %5, %2 ; <i32> [#uses=1] br label %flop.exit I thought that tail call means dropping current stack (i.e. return will be to the upper frame, so next instruction should be ret %5), but according to this code it will do mul for it. And in native assembly there is simple call without tail optimisation (even with appropriate flag for llc) Can sombody explain why clang generates such code? As well I can't understand why llvm have tail call if it can simply check that next ret will use result of prev call and later do appropriate optimisation or generate native equivalent of tail-call instruction?

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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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• C: Proper syntax for allocating memory using pointers to pointers.

  - by ~kero-05h

  This is my first time posting here, hopefully I will not make a fool of myself. I am trying to use a function to allocate memory to a pointer, copy text to the buffer, and then change a character. I keep getting a segfault and have tried looking up the answer, my syntax is probably wrong, I could use some enlightenment. /* My objective is to pass a buffer to my Copy function, allocate room, and copy text to it. Then I want to modify the text and print it.*/ #include <stdio.h> #include <stdlib.h> #include <string.h> int Copy(char **Buffer, char *Text); int main() { char *Text = malloc(sizeof(char) * 100); char *Buffer; strncpy(Text, "1234567890\n", 100); Copy(&Buffer, Text); } int Copy(char **Buffer, char *Text) { int count; count = strlen(Text)+1; *Buffer = malloc(sizeof(char) * count); strncpy(*Buffer, Text, 5); *Buffer[2] = 'A'; /* This results in a segfault. "*Buffer[1] = 'A';" results in no differece in the output. */ printf("%s\n", *Buffer); }

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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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• why i^=j^=i^=j isn't equal to *i^=*j^=*i^=*j

  - by klvoek

  In c , when there is variables (assume both as int) i less than j , we can use the equation i^=j^=i^=j to exchange the value of the two variables. For example, let int i = 3, j = 5; after computed i^=j^=i^=j, I got i = 5, j = 3 . What is so amazing to me. But, if i use two int pointers to re-do this , with *i^=*j^=*i^=*j , use the example above what i got will be i = 0 and j = 3. Then, describe it simply: In C 1 int i=3, j=5; i^=j^=i^=j; // after this i = 5, j=3 2 int i = 3, j= 5; int *pi = &i, *pj = &j; *pi^=*pj^=*pi^=*pj; // after this, $pi = 0, *pj = 5 In JavaScript var i=3, j=5; i^=j^=i^=j; // after this, i = 0, j= 3 the result in JavaScript makes this more interesting to me my sample code , on ubuntu server 11.0 & gcc #include <stdio.h> int main(){ int i=7, j=9; int *pi=&i, *pj=&j; i^=j^=i^=j; printf("i=%d j=%d\n", i, j); i=7, j==9; *pi^=*pj^=*pi^=*pj printf("i=%d j=%d\n", *pi, *pj); } however, i had spent hours to test and find out why, but nothing means. So, please help me. Or, just only i made some mistake???

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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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• Using libgrib2c in c++ application, linker error "Undefined reference to..."

  - by Rich

  EDIT: If you're going to be doing things with GRIB files I would recommend the GDAL library which is backed by the Open Source Geospatial Foundation. You will save yourself a lot of headache :) I'm using Qt creator in Ubuntu creating a c++ app. I am attempting to use an external lib, libgrib2c.a, that has a header grib2.h. Everything compiles, but when it tries to link I get the error: undefined reference to 'seekgb(_IO_FILE*, long, long, long*, long*) I have tried wrapping the header file with: extern "C"{ #include "grib2.h" } But it didn't fix anything so I figured that was not my problem. In the .pro file I have the line: include($${ROOT}/Shared/common/commonLibs.pri) and in commonLibs.pri I have: INCLUDEPATH+=$${ROOT}/external_libs/g2clib/include LIBS+=-L$${ROOT}/external_libs/g2clib/lib LIBS+=-lgrib2c I am not encountering an error finding the library. If I do a nm command on the libgrib2c.a I get: nm libgrib2c.a | grep seekgb seekgb.o: 00000000 T seekgb And when I run qmake with the additional argument of LIBS+=-Wl,--verbose I can find the lib file in the output: attempt to open /usr/lib/libgrib2c.so failed attempt to open /usr/lib/libgrib2c.a failed attempt to open /mnt/sdb1/ESMF/App/ESMF_App/../external_libs/linux/qwt_6.0.2/lib/libgrib2c.so failed attempt to open /mnt/sdb1/ESMF/App/ESMF_App/../external_libs/linux/qwt_6.0.2/lib/libgrib2c.a failed attempt to open ..//Shared/Config/lib/libgrib2c.so failed attempt to open ..//Shared/Config/lib/libgrib2c.a failed attempt to open ..//external_libs/libssh2/lib/libgrib2c.so failed attempt to open ..//external_libs/libssh2/lib/libgrib2c.a failed attempt to open ..//external_libs/openssl/lib/libgrib2c.so failed attempt to open ..//external_libs/openssl/lib/libgrib2c.a failed attempt to open ..//external_libs/g2clib/lib/libgrib2c.so failed attempt to open ..//external_libs/g2clib/lib/libgrib2c.a succeeded Although it doesn't show any of the .o files in the library is this because it is a c library in my c++ app? in the .cpp file that I am trying to use the library I have: #include "gribreader.h" #include <stdio.h> #include <stdlib.h> #include <external_libs/g2clib/include/grib2.h> #include <Shared/logging/Logger.hpp> //------------------------------------------------------------------------------ /// Opens a GRIB file from disk. /// /// This function opens the grib file and searches through it for how many GRIB /// messages are contained as well as their starting locations. /// /// \param a_filePath. The path to the file to be opened. /// \return True if successful, false if not. //------------------------------------------------------------------------------ bool GRIBReader::OpenGRIB(std::string a_filePath) { LOG(notification)<<"Attempting to open grib file: "<< a_filePath; if(isOpen()) { CloseGRIB(); } m_filePath = a_filePath; m_filePtr = fopen(a_filePath.c_str(), "r"); if(m_filePtr == NULL) { LOG(error)<<"Unable to open file: " << a_filePath; return false; } LOG(notification)<<"Successfully opened GRIB file"; g2int currentMessageSize(1); g2int seekPosition(0); g2int lengthToBeginningOfGrib(0); g2int seekLength(32000); int i(0); int iterationLimit(300); m_GRIBMessageLocations.clear(); m_GRIBMessageSizes.clear(); while(i < iterationLimit) { seekgb(m_filePtr, seekPosition, seekLength, &lengthToBeginningOfGrib, &currentMessageSize); if(currentMessageSize != 0) { LOG(verbose) << "Adding GRIB message location " << lengthToBeginningOfGrib << " with length " << currentMessageSize; m_GRIBMessageLocations.push_back(lengthToBeginningOfGrib); m_GRIBMessageSizes.push_back(currentMessageSize); seekPosition = lengthToBeginningOfGrib + currentMessageSize; LOG(verbose) << "GRIB seek position moved to " << seekPosition; } else { LOG(notification)<<"End of GRIB file found, after "<< i << " GRIB messages."; break; } } if(i >= iterationLimit) { LOG(warning) << "The iteration limit of " << iterationLimit << "was reached while searching for GRIB messages"; } return true; } And the header grib2.h is as follows: #ifndef _grib2_H #define _grib2_H #include<stdio.h> #define G2_VERSION "g2clib-1.4.0" #ifdef __64BIT__ typedef int g2int; typedef unsigned int g2intu; #else typedef long g2int; typedef unsigned long g2intu; #endif typedef float g2float; struct gtemplate { g2int type; /* 3=Grid Defintion Template. */ /* 4=Product Defintion Template. */ /* 5=Data Representation Template. */ g2int num; /* template number. */ g2int maplen; /* number of entries in the static part */ /* of the template. */ g2int *map; /* num of octets of each entry in the */ /* static part of the template. */ g2int needext; /* indicates whether or not the template needs */ /* to be extended. */ g2int extlen; /* number of entries in the template extension. */ g2int *ext; /* num of octets of each entry in the extension */ /* part of the template. */ }; typedef struct gtemplate gtemplate; struct gribfield { g2int version,discipline; g2int *idsect; g2int idsectlen; unsigned char *local; g2int locallen; g2int ifldnum; g2int griddef,ngrdpts; g2int numoct_opt,interp_opt,num_opt; g2int *list_opt; g2int igdtnum,igdtlen; g2int *igdtmpl; g2int ipdtnum,ipdtlen; g2int *ipdtmpl; g2int num_coord; g2float *coord_list; g2int ndpts,idrtnum,idrtlen; g2int *idrtmpl; g2int unpacked; g2int expanded; g2int ibmap; g2int *bmap; g2float *fld; }; typedef struct gribfield gribfield; /* Prototypes for unpacking API */ void seekgb(FILE *,g2int ,g2int ,g2int *,g2int *); g2int g2_info(unsigned char *,g2int *,g2int *,g2int *,g2int *); g2int g2_getfld(unsigned char *,g2int ,g2int ,g2int ,gribfield **); void g2_free(gribfield *); /* Prototypes for packing API */ g2int g2_create(unsigned char *,g2int *,g2int *); g2int g2_addlocal(unsigned char *,unsigned char *,g2int ); g2int g2_addgrid(unsigned char *,g2int *,g2int *,g2int *,g2int ); g2int g2_addfield(unsigned char *,g2int ,g2int *, g2float *,g2int ,g2int ,g2int *, g2float *,g2int ,g2int ,g2int *); g2int g2_gribend(unsigned char *); /* Prototypes for supporting routines */ extern double int_power(double, g2int ); extern void mkieee(g2float *,g2int *,g2int); void rdieee(g2int *,g2float *,g2int ); extern gtemplate *getpdstemplate(g2int); extern gtemplate *extpdstemplate(g2int,g2int *); extern gtemplate *getdrstemplate(g2int); extern gtemplate *extdrstemplate(g2int,g2int *); extern gtemplate *getgridtemplate(g2int); extern gtemplate *extgridtemplate(g2int,g2int *); extern void simpack(g2float *,g2int,g2int *,unsigned char *,g2int *); extern void compack(g2float *,g2int,g2int,g2int *,unsigned char *,g2int *); void misspack(g2float *,g2int ,g2int ,g2int *, unsigned char *, g2int *); void gbit(unsigned char *,g2int *,g2int ,g2int ); void sbit(unsigned char *,g2int *,g2int ,g2int ); void gbits(unsigned char *,g2int *,g2int ,g2int ,g2int ,g2int ); void sbits(unsigned char *,g2int *,g2int ,g2int ,g2int ,g2int ); int pack_gp(g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *, g2int *); #endif /* _grib2_H */ I have been scratching my head for two days on this. If anyone has an idea on what to do or can point me in some sort of direction, I'm stumped. Also, if you have any comments on how I can improve this post I'd love to hear them, kinda new at this posting thing. Usually I'm able to find an answer in the vast stores of knowledge already contained on the web.

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• C to Assembly code - what does it mean

  - by Smith

  I'm trying to figure out exactly what is going on with the following assembly code. Can someone go down line by line and explain what is happening? I input what I think is happening (see comments) but need clarification. .file "testcalc.c" .section .rodata.str1.1,"aMS",@progbits,1 .LC0: .string "x=%d, y=%d, z=%d, result=%d\n" .text .globl main .type main, @function main: leal 4(%esp), %ecx // establish stack frame andl $-16, %esp // decrement %esp by 16, align stack pushl -4(%ecx) // push original stack pointer pushl %ebp // save base pointer movl %esp, %ebp // establish stack frame pushl %ecx // save to ecx subl $36, %esp // alloc 36 bytes for local vars movl $11, 8(%esp) // store 11 in z movl $6, 4(%esp) // store 6 in y movl $2, (%esp) // store 2 in x call calc // function call to calc movl %eax, 20(%esp) // %esp + 20 into %eax movl $11, 16(%esp) // WHAT movl $6, 12(%esp) // WHAT movl $2, 8(%esp) // WHAT movl $.LC0, 4(%esp) // WHAT?!?! movl $1, (%esp) // move result into address of %esp call __printf_chk // call printf function addl $36, %esp // WHAT? popl %ecx popl %ebp leal -4(%ecx), %esp ret .size main, .-main .ident "GCC: (Ubuntu 4.3.3-5ubuntu4) 4.3.3" .section .note.GNU-stack,"",@progbits Original code: #include <stdio.h> int calc(int x, int y, int z); int main() { int x = 2; int y = 6; int z = 11; int result; result = calc(x,y,z); printf("x=%d, y=%d, z=%d, result=%d\n",x,y,z,result); }

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• Exercise for exam about comparing string on file

  - by Telmo Vaz

  I'm trying to do this exercise for my exam tomorrow. I need to compare a string of my own input and see if that string is appearing on the file. This needs to be done directly on the file, so I cannot extract the string to my program and compare them "indirectly". I found this way but I'm not getting it right, and I don't know why. The algorithm sounds good to me. Any help, please? I really need to focus on this one. Thanks in advance, guys. #include<stdio.h> void comp(); int main(void) { comp(); return 0; } void comp() { FILE *file = fopen("e1.txt", "r+"); if(!file) { printf("Not possible to open the file"); return; } char src[50], ch; short i, len; fprintf(stdout, "What are you looking for? \nwrite: "); fgets(src, 200, stdin); len = strlen(src); while((ch = fgetc(file)) != EOF) { i = 0; while(ch == src[i]) { if(i <= len) { printf("%c - %c", ch, src[i]); fseek(file, 0, SEEK_CUR + 1); i++; } else break; } } }

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• c++ Mixing printf with wprintf (or cout with wcout)

  - by Bo Jensen

  I know you should not mix printing with printf,cout and wprintf,wcout, but have a hard time finding a good answer why and if it is possible to get round it. The problem is I use a external library that prints with printf and my own uses wcout. If I do a simple example it works fine, but from my full application it simply does not print the printf statements. If this is really a limitation, then there would be many libraries out there which can not work together with wide printing applications. Any insight on this is more than welcome. Update : I boiled it down to : #include <stdio.h> #include <stdlib.h> #include <iostream> #include <readline/readline.h> #include <readline/history.h> int main() { char *buf; std::wcout << std::endl; /* ADDING THIS LINE MAKES PRINTF VANISH!!! */ rl_bind_key('\t',rl_abort);//disable auto-complete while((buf = readline("my-command : "))!=NULL) { if (strcmp(buf,"quit")==0) break; std::wcout<<buf<< std::endl; if (buf[0]!=0) add_history(buf); } free(buf); return 0; } So I guess it might be a flushing problem, but it still looks strange to me, I have to check up on it.

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• Inorder tree traversal in binary tree in C

  - by srk

  In the below code, I'am creating a binary tree using insert function and trying to display the inserted elements using inorder function which follows the logic of In-order traversal.When I run it, numbers are getting inserted but when I try the inorder function( input 3), the program continues for next input without displaying anything. I guess there might be a logical error.Please help me clear it. Thanks in advance... #include<stdio.h> #include<stdlib.h> int i; typedef struct ll { int data; struct ll *left; struct ll *right; } node; node *root1=NULL; // the root node void insert(node *root,int n) { if(root==NULL) //for the first(root) node { root=(node *)malloc(sizeof(node)); root->data=n; root->right=NULL; root->left=NULL; } else { if(n<(root->data)) { root->left=(node *)malloc(sizeof(node)); insert(root->left,n); } else if(n>(root->data)) { root->right=(node *)malloc(sizeof(node)); insert(root->right,n); } else { root->data=n; } } } void inorder(node *root) { if(root!=NULL) { inorder(root->left); printf("%d ",root->data); inorder(root->right); } } main() { int n,choice=1; while(choice!=0) { printf("Enter choice--- 1 for insert, 3 for inorder and 0 for exit\n"); scanf("%d",&choice); switch(choice) { case 1: printf("Enter number to be inserted\n"); scanf("%d",&n); insert(root1,n); break; case 3: inorder(root1); break; default: break; } } }

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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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• 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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