[SOLVED] Why do programs with arguments passed in argc and argv get different results when executed in different ways

Issue

void main(int argc,char *argv[])
{
     for (int i = 0; i < argc; i++)
     {
         printf("%s ", argv[i]);
     }
}

when I use command ./test 1 2 3 in terminal to execute this program, I got result ./test 1 2 3 ,but when I use function execl("/usr/src/test", "1", "2", "3", NULL) in another program I got result 1 2 3,why?

Solution

The syntax of execl() is:

int execl(const char *path, const char *arg0, ..., /*, (char *)0, */);

So you have

path = "/usr/src/test"
arg0 = "1"
arg1 = "2"
arg3 = "3"

The argN parameters are put into the argv array of the new process.

You have to repeat the path as arg0 to put that into argv[0].

execl("/usr/src/test", "/usr/src/test", "1", "2", "3", NULL)

This isn't done automatically because argv[0] isn't required to be the same as the program path, and there are some situations where it isn't (for instance, login shells are invoked by adding a - prefix in argv[0]).

Answered By - Barmar

Answer Checked By - Marie Seifert (WPSolving Admin)