Why does this program prints an empty line? - c

Question

This program is called program.c. When I run ./program echo test, I would expect the program to print test, even if the command is run in a subshell. Why is the output an empty line? Does it have to do with filepath? When I try ./program /bin/echo test I still get an empty line output.
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
int function(char **argv) {
execl("/bin/bash", "sh", "-c", *argv, argv, (char*)NULL);
}
int main(int argc, char **argv) {
int return2;
function(argv + 1);
}

Answer 1

There are two problems with your program.
The first problem is that the argument argv to execl() does not do what you think it does. The pointer argv is a pointer to a char * (i.e. a char **), and it is only permitted to pass a char * as an argument to execl(). That is, variable argument lists in C do not work how you are expecting them to.
In order to accomplish what you want, consider not using execl() and use execv() instead, passing an array of char *s which you have constructed yourself. You could do this in the following way, for example:
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
int function(char **argv) {
// Consider passing `argc` as well to eliminate
// the need for this loop.
int len = 0;
while(argv[len]) {
len++;
}
char *new_argv[len + 3];
new_argv[0] = "sh";
new_argv[1] = "-c";
for(int i = 0; i <= len; i++) {
new_argv[i + 2] = argv[i];
}
execv("/bin/bash", new_argv);
}
int main(int argc, char **argv) {
function(argv+1);
}
However, you still have a problem: ./program echo test will still print a blank line because if you execute sh -c echo test in a terminal you just get a blank line! In order to fix this, you need to do ./program 'echo test' (corresponding to sh -c 'echo test' at the terminal), which should then work.