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]

Wayne Shu wrote:

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
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[James Kanze]

On Jul 29, 2:58 pm, Wayne Shu <Wayne...@gmail .comwrote:

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.

--
James Kanze (Gabi Software) email: ja*********@gma il.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]