Question

I know this is fairly basic, but I’m still stuck.
So I have a function that needs to take in a variable n, so this is my main function

int main(int argc, char* argv){
  sort(argv[1]);
    }

And I’m calling the program like this:

    ./sort 4 <text.txt

But the number 4 doesnt get recognized or passed into the function. What am I doing wrong? I know that argv[0] should hold the name of program itself and each one from there on should hold the arguments.

Answer 1

You should try to print them all.

#include <stdio.h>

int main(int argc, const char *argv[])
{
    int i = 0;
    for (; i < argc; ++i) {
        printf("argv[%d] = '%s'\n", i, argv[i]);
    }
    return 0;
}

Running that code with ./a.out 4 < /somefile gives me:

argv[0] = './a.out'
argv[1] = '4'

Eventually you’ll have to remember that the ‘4’ is a pointer to an array of characters, and you might have to parse it into an integer.