Trying to use execvp() in C with user input in unix

Question

I'm trying to make a program that will prompt the user for a command, then use exec to execute that command.

For instance if they gave me "ls -la" I would have to execute that command. I've tried the following code:

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

int main()
{

    int ret, num_args;

    printf("Enter number of arguments (Example: \"ls -la\" has 1 argument): ");
    scanf("%d", &num_args);

    char *cmd[num_args];

    printf("Enter command name: ");
    scanf("%s", &cmd[0]);

    int i;
    for (i = 0; i < num_args; i++)
    {
            printf("Enter parameter: ");
            scanf("%s", &cmd[i]);
    }

    execvp(cmd[0], cmd);
}

However, when I tried the following run it gave me a "segmentation fault"

$ ./a.out 
Enter number of arguments (Example: "ls -la" has 1 argument): 2
Enter command name: ls
Enter parameter: -la
Enter parameter: .
Segmentation fault
$

Any ideas?

Answer 1

You need to allocate memory for your strings. The following line only allocates num_args worth of pointers to char:

char *cmd[num_args];

First of all, you'll be getting num_args + 1 strings (don't forget that the command itself is cmd[0]). The easiest way is to statically allocate the memory as an array of character buffers:

const unsigned int MAX_LEN = 512; // Arbitrary number
char cmd[num_args + 1][MAX_LEN];

However, now you can't use scanf to read in a line because the user could input a string that's longer than your character buffer. Instead, you'll have to use fgets, which can limit the number of characters the user can input:

fgets(cmd[i], MAX_LEN, stdin);

Keep in mind that fgets also reads newline characters, so make sure to strip any stray ones that show up (but don't assume that they're there).