CODEWITHSUNDEEP
c
Philip
Philip

Reputation: 5917

Type-qualifiers on function return type

Given the following C source code:

const int foo(void)
{
    return 42;
}

gcc compiles without errors, but with -Wextra or -Wignored-qualifiers, the following warning appears:

warning: type qualifiers ignored on function return type

I understand that there's good reason in C++ to distinguish between const functions and non-const functions, e.g. in the context of operator overloading.

In plain C however, I fail to see why gcc doesn't emit an error, or more concisely, why the standard allows const functions.

Why is it allowed to use type qualifiers on function return types?

Upvotes: 7

Views: 10712

Answers (3)

unwind
unwind

Reputation: 399813

Because it makes sense for pointer types, my guess is that it simply keeps the grammar simpler, so nobody thinks it's worth making const non-pointer return values be an error.

Also, your comparison with C++ is a bit off, since a constant method in C++ is declared by having const last:

int foo() const;

There is no relationship between a method being const, and a method having a const return value, they are completely distinct things. The syntax makes this fairly clear.

Upvotes: 2

hmjd
hmjd

Reputation: 121971

Consider:

#include <stdio.h>

const char* f()
{
    return "hello";
}
int main()
{
    const char* c = f();

    *(c + 1) = 'a';

    return 0;
}

If const were not permitted on the return value then the code would compile (and cause undefined behaviour at runtime).

const is useful when a function returns a pointer to something unmodifiable.

Upvotes: 3

pmg
pmg

Reputation: 108978

It's irrelevant if the value returned from the function is qualified as const.

You cannot change the value returned even if it wasn't qualified.

foo() = -42; /* impossible to change the returned value */

So using const is redundant (and normally omitted).

Upvotes: 1

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