Deleting overloaded function. C++11. Call of overloaded ... is ambiguous

Question

There is global function (just example):

void func( int i )
{
    std::cout << i + 100 << std::endl;
}

I assume that calling this function with char argument does not make any sense so I use delete:

void func(char) = delete;

So I expect that following calls should be possible:

func(1);
func(3.1);
func(true);

And call with char argument should be forbiden:

func('a');

But that is not true. When calling func('a') I get as expected:

error: use of deleted function ‘void func(char)’

But during calling func(2.3) I get:

error: call of overloaded ‘func(double)’ is ambiguous

Why do I get this error? Without deleting function with char arguments double was converted to int and func(int) was called, why now it is forbidden?

"Why do I get this error?" You are aware that char can be automatically converted to an int and to a double in turn? — user0042
Everything you have after EDITED seems like an answer to your question. If that's your answer, please remove it from the question and post it as an answer. If, as it seems, you're just echoing an existing one, don't post a duplicate answer; just delete it from your question. — Nic

vsoftco vsoftco · Accepted Answer · 2017-12-28T17:44:21

When you call

func(2.3)

you pass a double to the function. The list of candidates contains both func(int) and func(char), as overload resolution happens before delete kicks in:

If the function is overloaded, overload resolution takes place first, and the program is only ill-formed if the deleted function was selected _{Ref: cppreference.com, see Avishai's answer for precise standard quotes.}

Now double can be converted both to char and int, hence the ambiguity.

Without deleting function with char arguments double was converted to int and func(int) was called, why now it is forbidden?

You get an ambiguity error even without deleteing the char version, see live here. Of course if you only define the func(int), then there will be no ambiguity so double will happily be converted to an int.

Deleting overloaded function. C++11. Call of overloaded ... is ambiguous

3 Answers