重载决策中的const指针

Question

GCC treats these two function declarations as equivalent:

void F(int* a) { }
void F(int* const a) { }

test.cpp: In function 'void F(int*)':

test.cpp:235: error: redefinition of 'void F(int*)'

test.cpp:234: error: 'void F(int*)' previously defined here

This makes some sense because a caller will always ignore the const in this case... it only affects the usage of the parameter 'a' inside of the function.

What I'm wondering is where (if anywhere) the standard says that it's specifically OK to discard qualifiers on pointers used as function arguments for the purpose of overload resolution.

(My real issue is that I'd like to figure out where GCC strips these pointless qualifiers internally, and since the C++ frontend of GCC is littered with comments referencing the standard, the relevant section of the standard might help me find the correct spot.)

Answer 1

Standard says in 8.3.5/3 that for the purposes of determining the function type any cv-qualifiers that directly qualify the parameter type are deleted. Ie it literally says that a function declared as

void foo(int *const a);

has function type void (int *) .

A pedantic person might argue that this is not conclusive enough to claim that the above declaration should match the definition like this one

void foo(int *a)
{
}

or that it should make the code with dual declaration (as in your example) ill-formed, since neither of these concepts are described in the standard in terms of function types .

I mean, we all know that these const were intended to be ignored for all external purposes, but so far I was unable to find the wording in the standard that would conclusively state exactly that. Maybe I missed something.

Actually, in 13.1/3 it has a "Note" that says that function declarations with equivalent parameter declarations (as defined in 8.3.5) declare the same function . But it is just a note, it is non-normative, which suggests that somewhere in the standard there should be some normative text on the same issue.

Answer 2

I think it is basically as prohibited as this:

void foo(int a) {}
void foo(const int a) {}

const on non-references doesn't participate in overloading.

In fact you could even declare

void foo(int a);

and later define

void foo(const int a) {}

where the constness is purely an implementation detail which the caller doesn't care about.

Answer 3

It's the same as:

void foo(int);
void foo(const int);

Being the same to the caller. This is because the function is getting a copy by-value no matter what, so the caller doesn't care if it's thought of as const or not; it makes no difference to it.

It's not legal for the compiler to ignore such things, but there is no difference in overload resolution. The const applies to the implementation of the function.

Illegal would be if the compiler treated:

void foo(int i)
{
    i = 5; // good
}

void foo(const int)
{
    i = 5; // lolwut?
}

The same, by ignoring the const .

Answer 4

I believe it's the other way around. Any pointer, even nonconst, can be treated like const :).