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Compiling c code programatically in linux terminal gcc

由 杀马特。学长 韩版系。学妹 提交于 2019-12-06 04:38:06

I'm not completely following your logic, put here are some possible problems:

First of all, where do you get the path from; this program may be simpler using command-line arguments than it is how you are doing it.

This means, you declare main as follows in your main file: int main( int argc, char *argv[] )

In this form, argc is the number of arguments passed INCLUDING the file name, and argv[0] is a string holding your file's name.

So, rather than rely on the program to generate the file for you, you do it somewhat like this:

#include "stdlib.h"
#include "stdio.h"
#include "string.h"
int main( int argc, char *argv[] ) {
  if (argc != 3) {
    fprintf( stderr, "Usage: %s %s\n", argv[0], "input_file output_file" );
    fprintf( stderr, "Do not add an extension to output_file.\n" );
    return EXIT_FAILURE;
  }
  unsigned int i = 0;
  for (i = 0; i < strlen( argv[2] ); i++) {
    if (argv[2][i] == '\n' || argv[2][i] == '\0') {
      argv[2][i] = '\0';
      break;
    }
  }
  char format_string[150];
  sprintf( format_string, "gcc.exe -Wall -Wextra -c %s -o %s.o", argv[1], argv[2] );
  char format_string2[150];
  sprintf( format_string2, "gcc.exe %s -o %s.exe", argv[1], argv[2] );
  system( format_string );
  system( format_string2 );
  return EXIT_SUCCESS;
}

With the above program, you just create the input file then pass the inpu file as the first argument and your output to the second one. This file does make two assumptions: input_file exists and the path to gcc is in your enviroment's path variable.

Finally, don't forget that main does NOT need a prototype.

标签
c
Linux
gcc
compilation
terminal
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