Question

Assume I have char **argv. How do I determine its size?

I have a string – an example would be: sleep 30 & that is held in argv. I would like to be able to access the last array in *argv. In this case, the last array contains &. How can I access it? strlen(argv) doesn’t seem to work properly. sizeof() obviously wouldn’t work properly because **argv is a pointer.

Note: I am not talking about **argv as an argument in main(), therefore, I do not have argc or any other indicator of how long the string is.

Answer 1

EDIT: Edited to work with a custom array of strings. A NULL pointer indicates the end of the array. Although this declares an array of 4 strings, this method could be used with a dynamically sized array.

#include <stdio.h>

int main()
{
    char* custom[4] = { "sleep", "30", "&", NULL };
    int last;
    for (last = 0; custom[last + 1]; last++);
    printf("%i - %s\n", last, custom[last]);
    return 0;
}

// ./a.out
// > 2 - &

For this to work for you, you would have to edit your program to explicitly include an extra NULL string in your char** when you build it. Without that indicator, the address after the last string wouldn’t necessarily be NULL, so you could include garbage in the count or cause a segmentation fault.