Re: GCC behave different for cv-qualifier function.

[zhang qingshan <steven dot zhang54373 at gmail dot com>]

On Wed, Dec 15, 2010 at 10:57 AM, Jonathan Wakely <jwakely.gcc@gmail.com> wrote:
> Please reply on the mailing list, I don't want to give you private C++
> lessons. Thanks.
>
> On 15 December 2010 02:20, zhang qingshan wrote:
>> On Tue, Dec 14, 2010 at 5:11 PM, Jonathan Wakely wrote:
>>> On 14 December 2010 07:26, zhang qingshan wrote:
>>>> std 8.3.2/1 says
>>>>
>>>> Cv-qualified references are ill-formed except when the cv-qualifiers
>>>> are introduced through
>>>> the use of a typedef (7.1.3) or of a template type argument (14.3), in
>>>> which case the cv-qualifiers are ignored
>>>>
>>>> following code is tested at GCC 4.5.0.
>>>>
>>>> template <typename T>
>>>> void fun(const T&t);
>>>> void foo();
>>>> void Test() {
>>>> ?fun(foo);
>>>> }
>>>>
>>>> It seems that, fun should be resolved as void (void (&)()), as const
>>>> is ignore here.
>>>
>>> No.
>
> Const is not ignored here because of 8.3.2/1, because you do not have
> a const-qualified reference.
>
> Please read the FAQs I sent you, maybe you are still confused about
> what "const T&" means.
>
Ok, all my examples are not relative with qulifier reference. In fact,
I can fully understand the difference between const T& and T&const.
It acts the same sematic as const T* and T*const, AFAIK.

Sadly, I cannot understand what the situation would be if T is
replaced by function.

When the template argument substitudition happens, T --> void (),
const T --> void (), as it is a const qualifier function, which const
is ignore.
finally, reference apply for this type descriptor. const T& -->
void(&)(). I believe that, I have missed something here.

>> if it is resolved as void(void(const&)()), I cannot compile the
>> following code with GCC.
>> void fun(void (const &f)())
>> {
>> }
>
> That syntax is not valid.
>
> Maybe you want:
>
> typedef void (func_type)();
>
> void fun (const func_type& f)
> {
> }
>
> That would be a reference to a const function, but see 8.3.5/5 - there
> are no cv-qualified function types, so the const is ignored. ?But
> that's a different rule to the one you keep quoting from 8.3.2/1.
>
> Const is ignored on references, but that's not what your example
> shows. Your example adds a const to a function type. Your example is
> not related to 8.3.2/1
>
> In the error message you got, it shows "void (const&)()" but that is
> NOT a const-qualified reference, it's a const-qualified function, so
> the const is ignored because of 8.3.5/5. ?You can demonstrate that by
> providing a specialization without the const, which resolves the link
> error:
>
> template <typename T>
> void fun(const T&t);
>
> typedef void(func_type)();
> template<>
> void fun<func_type>(func_type&) { }
>
> void foo();
> void Test() {
> ? fun(foo);
> }
>
> This shows that "void (const&)()" and "void (&)()" are equivalent.
> That's still nothing to do with 8.3.2/1
>
> THIS would be a const qualified reference introduced by a template
> type argument, so the const would be ignored:
>
> template<typename T>
> void f(const T);
>
> void Test() {
> ? int i = 0;
> ? f<int&>(i);
> }
>
> Here the template parameter is "int&" so the function argument type is
> "int& const" but that forms a cv-qualified reference, so the const is
> ignored, and the type of the template function specialization is
> "f<int&>(int&)"

Thanks for your example and detail explain about the cv-qualified
reference and cv-qualified function.