c - Const on print of mutable object

Question

Is it necessary to declare a mutable object const if the function does not modify it, or is this a redundancy that is not required? For example:

#include<stdio.h>

void print_num_ptr(const int *n) {
    printf("Example print: %d", *n);
}
int main(void) {
    int i=4;
    int *ptr_i=&i;
    print_num_ptr((const int*)ptr_i);
    (*ptr_i) ++;
    print_num_ptr((const int*)ptr_i);

}

Vs.

#include<stdio.h>

void print_num_ptr(int *n) {
    printf("Example print: %d", *n);
}
int main(void) {
    int i=4;
    int *ptr_i=&i;
    print_num_ptr(ptr_i);
    (*ptr_i) ++;
    print_num_ptr(ptr_i);
}

The first one just seems like a lot of headache unless there's a reason to use it over the first.

question from:https://stackoverflow.com/questions/66056049/const-on-print-of-mutable-object

Answer 1

It's a good habit to declare unchanging arguments as const both to advertise in the function signature that you will not change them, and to allow the compiler to optimize around that parameter.

Any function with a non-const argument implies it can or will modify that value, which if not true, communicates your intentions incorrectly.

Adding const to everything might seem like a useless ritual, but it has a very important function.

As a note, it's usually not necessary to cast to a const version of same. The compiler does that for you automatically. const acts as a safety that's implicitly applied, but won't be removed without going out of your way to break it off.