Re: [boost] decltype and incomplete types

Subject: Re: [boost] decltype and incomplete types
From: Daniel Walker (daniel.j.walker_at_[hidden])
Date: 2010-04-08 16:53:01

On Thu, Apr 8, 2010 at 2:18 PM, Eric Niebler <eric_at_[hidden]> wrote:
> On 4/7/2010 1:04 PM, Daniel Walker wrote:
>>
>> On Wed, Apr 7, 2010 at 12:46 PM, Eric Niebler<eric_at_[hidden]>  wrote:
>>>
>>> On 4/7/2010 9:21 AM, Daniel Walker wrote:
>>>>
>>>> After considering this for a while, I believe there is an argument in
>>>> support of the current decltype specification, at least with respect
>>>> to result_of. result_of is supposed to give the type of the result of
>>>> a function call, i.e. the type of a function call expression. A call
>>>> expression cannot have an incomplete type. So, result_of cannot have
>>>> an incomplete type.
>>>
>>> <snip>
>>>
>>> Strongly disagree. Decltype (decl-type) yields the "declared type" of an
>>> expression. Consider:
>>
>> Not to be pedantic, but call expressions don't exactly have a single
>> "declared type." Their type is determined dynamically by the compiler
>> according to the context in which they are evaluated.   The same call
>> expression may have different types in different context.
>>
>>>
>>>  // found in some header somewhere
>>>  struct S;
>>>  S foo();
>>>
>>> What is the "declared type" of the expression "foo()"? Obviously, it is
>>> S.
>>
>> It might seem so at first, but consider:
>>
>> struct S;
>> S foo(int);
>> int foo(S);
>>
>> The type of the call expression foo(x) depends on which function is
>> statically chosen after overload resolution. Overload resolution may
>> not succeed unless/until S is complete.
>
>
> 7.1.6.2/4:
> "The type denoted by decltype(e) is defined as follows:
> — if e is ... or if e names a set of overloaded functions, the program is
> ill-formed;"
>
> Note that your program is ill-formed if you try to take the decltype of an
> overloaded function set. It continues ...
>
> "— otherwise, if e is a function call (5.2.2) or an invocation of an
> overloaded operator (parentheses around e are ignored), decltype(e) is the
> return type of the statically chosen function"
>
> <snip the rest, which is not relevant>

If overload resolution succeeds then e names a statically chosen
function not an overload set. If overload resolution fails then e
names an overload set and the program is ill-formed. That's my
understanding at least. So, yes, you can use decltype on an overloaded
function and the program is not ill-formed, so long as overload
resolution succeeds. I believe that's true of call expressions in any
context.

Daniel Walker