Passing variables into c++ using the command line

[KazerTheKeen]

So I have this C++ code here and it compiles fine, but I was wondering how to pass in a variable for argv. Currently I'm running the code as such,  

"   g++ counter.cc #0 '21'   "

This compiles but it gets 0 as count. So what do I need to change? Ive tried using int count = atoi(&argv) and other similar variations but can't figure out where I'm going wrong with this.

On Linux btw if its relevant.

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

int main(int argc, char *argv[ ]){
    
    int count = atoi( argv[0]);
    printf("Count :%i \n", count);

    for(int i = 1; i <= count; i++){
        printf("Process: %d %i \n", getpid(), i );
    }
    return(count);
    perror("Return Failed");
    exit(EXIT_FAILURE);    
}

 

[Mira Yurizaki]

You pass in arguments like

./a.out 1 2 3 4 5

 

Arguments are space-separated terms that fill in the argv array. Argc is used to tell you how many arguments there are. Also, important note, the first entry in argv is always the command. In the example above, argv[0] is ./a.out. So you want to make sure you have at least 2 arguments before proceeding.

[KazerTheKeen]

Ok, Thank you, Ive been struggling on that for far too long.

[Mira Yurizaki]

Also one more note since I've read into it a bit as a refresher, argv[argc] is always going to be a null pointer. So don't try to do anything with argv[argc].

[Spigie]

When the arguments you are receiving are relatively simple, then argc and argv should be fine, in a situation like this they are . However sometimes the arguments you want to receive will be more complex, the GNU C library (which can be used in C++ as well) has a header called argp, which is to help with parsing those longer arguments, an example of how to use it is here, however it is not directly a part of C++ so the syntax of argp is C-like, which can get complex and messy sometimes (but all code can get complex and messy).