I encounter a question while I am reading a textbook - Unix System Programming
How big is the argument array passed as the second argument to
execvp
when you executeexeccmd
of Program 3.5 with the following command line?execcmd ls -l *.c
Answer: The answer depends on the number of
.c
files in the current directory because the shell expands*.c
before passing the command line toexeccmd
.
Program 3.5:
#include <errno.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include "restart.h"
int main(int argc, char *argv[]) {
pid_t childpid;
if (argc < 2){ /* check for valid number of command-line arguments */
fprintf (stderr, "Usage: %s command arg1 arg2 ...\n", argv[0]);
return 1;
}
childpid = fork();
if (childpid == -1) {
perror("Failed to fork");
return 1;
}
if (childpid == 0) {
execvp(argv[1], &argv[1]);
perror("Child failed to execvp the command");
return 1;
}
if (childpid != r_wait(NULL)) {
perror("Parent failed to wait");
return 1;
}
return 0;
}
Why is the size of argument array passed depending on the number of .c files in the current directory? Isn't it the argument array just something like
argv[0] = "execcmd";
argv[1] = "ls";
argv[2] = "-l";
argv[3] = "*.c";
argv[4] = NULL;
Update: Find a link explains pretty well about the shell expansion. May useful for someone who see this post later also do not understand shell expansion. Description about shell expansion