Execv with arguments starting from second element

Question

I am making a program that can take another program as an argument and run it.

I am using execv() to run the second program, but how can I use argv[] (that comes from the main program, is a char*) as arguments for execv()?

The first element of argv[] is the name of the main program, so I need to start at the second element where the name of the next program is located.

I am thinking there might be different solutions, for example:

• Tell execv() to start from the second argument
• copy the content of argv[] (except first element) to another array and use that
• Something with pointers so maybe I can point to the second element and start there

What is the best way to do this? And can I get an example of how to do it?

Answer 1

First of all, the argv is actually a pointer to array of pointers. The more correct way to define it would be char **argv.

Defining it as *argv[] is just a syntactic sugar. So your third guess is a best one:

#include <stdio.h>

void print_array(int array_size, char **array) {
    for(int i=0; i<array_size; i++) {
        printf("\t%s\n", array[i]);
    }
}

int main(int argc, char **argv) {
    char **argv_new = argv+1;

    printf("argv:\n");
    print_array(argc, argv);
    printf("argv_new:\n");
    print_array(argc-1, argv_new);
    return 0;
}

You can also play with interchanging function arguments char **argv vs char *argv[] and you wont see a difference. But code like:

int main(int argc, char *argv[]) {
    char *argv_new[] = argv+1;

will give you an error - of invalid type conversion. That is due to a fact that defining an array on a stack requires to know the size of array during the compilation. But defining array as a function parameter - does not, the compiler in that case presumes that the stack manipulation is done in the caller function.