On 15 December 2010 08:50, zhang qingshan wrote: > > Sadly, I cannot understand what the situation would be if T is > replaced by function. That is governed by 8.3.5 not 8.3.2 > 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. I showed in my example that void (const&)() and void(&)() are equivalent. The reason you can't compile your other examples is only because you're using invalid syntax. C++ has very unhelpful declarator syntax for function types.
