some problems about exact match in the overload resolution.

[Wayne Shu]

Hi everyone.

In the following program, foo is an ambiguous call.

#include <iostream>
using namespace std;

void foo(int *);
void foo(int (&)[5]);

int main()
{
int arr[5] = {0, 1, 2, 3, 4};

foo(arr);

return 0;
}

why?
void foo(int (&)[5]); is an exact match
void foo(int (&)[5]); require an array to pointer conversion.

Regards.

 

[Victor Bazarov]

Hi everyone.

In the following program, foo is an ambiguous call.

#include <iostream>
using namespace std;

void foo(int *);
void foo(int (&)[5]);

int main()
{
int arr[5] = {0, 1, 2, 3, 4};

foo(arr);

return 0;
}

why?
void foo(int (&)[5]); is an exact match
void foo(int (&)[5]); require an array to pointer conversion.

You mean foo(int*) requires the conversion...

No, foo(int(&)[5]) is not an exact match. It requires binding
of a reference. An exact match would be a function with an array
as the argument, which is impossible.

V

 

[James Kanze]

In the following program, foo is an ambiguous call.

#include <iostream>
using namespace std;

void foo(int *);
void foo(int (&)[5]);

int main()
{
int arr[5] = {0, 1, 2, 3, 4};

foo(arr);

return 0;

}

why?
void foo(int (&)[5]); is an exact match

Actually, it's what the standard calls an lvalue transformation.

void foo(int (&)[5]); require an array to pointer conversion.

You mean "foo( int* )", of course. For purposes of overload
resolution, the standard considers it an lvalue transformation
as well; for purposes of ranking, both are considered exact
matches. There are additional considerations which can be taken
into account, but none apply here.