Understanding void and return 0/1 with an example

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#include <stdio.h>

void meow(int n);

int main(void)
{
    meow(3);
}

void meow(int n)
{
    for (int i = 0; i < n; i++)
    {
        printf("meow\n");
    }
}

[/dm_code_snippet]

 

In Lecture 1, along with the above code, it is mentioned:

“The void before the meow function means that it doesn’t return a value, and likewise in main we can’t do anything with the result of meow, so we just call it.”

I thought to make meow function return a value and so tried with this code:

[dm_code_snippet background=”no” background-mobile=”yes” slim=”yes” line-numbers=”yes” bg-color=”#abb8c3″ theme=”dark” language=”clike” wrapped=”yes” height=”” copy-text=”Copy Code” copy-confirmed=”Copied”]

#include <stdio.h>

int meow(int n);

int main(void)
{
    meow(3);
}

int meow(int n)
{
    for (int i = 0; i < n; i++)
    {
        printf("meow\n");
    
    }
    return 1;
}

[/dm_code_snippet]

This compiles successfully. I need to confirm if return 1 in the above code means that the program runs successfully. Or what is the utility of return 1? Can any function be made int type with bringing in return 0/return 1? Return 0 leads to non compiling in the above instance. My hunch (which actually perhaps not true) is suppose I make any error within the function meow and return 0 should lead to successful compilation as return 0 ensures that the program was unsuccessful in running.

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