execve() failing to launch program in C

问题

I am trying to spawn a new process using execve() from unistd.h on Linux. I have tried passing it the following parameters execve("/bin/ls", "/bin/ls", NULL); but get no result. I do not get an error either, the program just exits. Is there a reason why this is happening? I have tried launching it as root and regular user. The reason I need to use execve() is because I am trying to get it to work in an assembly call like so

program: db "/bin/ls",0

mov eax, 0xb
mov ebx, program
mov ecx, program
mov edx, 0
int 0x80

Thank you!

回答1:

The arguments that you're passing to execve are wrong. Both the second and third must be an array of char pointers with a NULL sentinel value, not a single pointer.

In other words, something like:

#include <unistd.h>
int main (void) {
    char * const argv[] = {"/bin/ls", NULL};
    char * const envp[] = {NULL};
    int rc = execve ("/bin/ls", argv, envp);
    return rc;
}

When I run that, I do indeed get a list of the files in the current directory.

回答2:

From the man pages,

int execve(const char *filename, char *const argv[], char *const envp[]);

So the problem in your case is that you haven't passed the 2nd and the 3rd argument correctly.

/* execve.c */

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

int
main(int argc, char *argv[])
{
    char *newargv[] = { NULL, "hello", "world", NULL };
    char *newenviron[] = { NULL };


newargv[0] = argv[1];

execve(argv[1], newargv, newenviron);


}
//This is a over-simplified version of the example in the man page

Run this as:

$ cc execve.c -o execve
$ ./execve ls

回答3:

Try reading man execve again. You are passing the wrong arguments to it. Pay particular attention to what the second argument should be.

Also, running your program under strace could be illuminating.

来源：https://stackoverflow.com/questions/8469818/execve-failing-to-launch-program-in-c