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Usage of argv[0]

int main(int argc, char* argv[])
{
   return 0;
}

Everybody recognizes this line. In some sources you can find that the argv[0] variable contains a full(?) path to the executable started including its name.

Recently I had noticed it is not always true. My statement is the argv[0] contains a string which was provided to the system to start the file (without following command line arguments), whatever it was, namely it can be a file name without an extenstion and path, a name and extension with no path, or a full path plus file name etc... Actually, argv[0] is a command line argument number 0, which can be anything sufficient to start an application and not necessary the full path to the file.

To prove my words you may do the following: compile this snippet.

#include "stdafx.h" //or #include "stdio.h"
#include "stdlib.h"

int main(int argc, char* argv[])
{
   printf("_pgmptr = %s\n", _pgmptr);
   printf("argv[0] = %s\n", argv[0]);
   printf("Hello World!\n");
   return 0;
}

Place it somewhere on you hard drive. Start cmd. Set current directory to the folder where the file is. Run it by name with extension. Run it by name without extention. Run it by entering full path to the file. Check the output. argv[0] is always different.

See that run-time global variable _pgmptr? For me it was always the same and equal to complete full path including name and extension... So, be aware what is going on under the hood even in such a simple thing.

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