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.
> 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&)"
