Decltype (* & fun) is weird?

Question

I have:

#include <type_traits> #include <stdio.h> void f() { printf("foo\n"); } int main() { printf("%d %d %d\n", std::is_same<decltype(*&f),decltype(f)>::value, std::is_function<decltype(*&f)>::value, std::is_function<decltype(f)>::value); (*&f)(); return 0; } 

what gives

 0 0 1 foo 

in g ++ 4.6.1 and 4.7.0.

Can someone explain this to me?

Answer 1

It is important to note that decltype has two meanings: it can be used to search for the declared type (hence its name) of an object or it can be used to test an expression. I use the entity here freely and do not mean a single term of the Standard, but, simply put, it can be a variable, function or (oddly enough, in my opinion) membership access. The type returned by checking the expression most often differs from the type of the expression itself, thus:

 int i; void foo(); struct { int i; } t; static_assert( std::is_same<decltype( i ), int>::value, "" ); static_assert( std::is_same<decltype( foo ), void()>::value, "" ); static_assert( std::is_same<decltype( ti ), int>::value, "" ); static_assert( std::is_same<decltype( (i) ), int&>::value, "" ); static_assert( std::is_same<decltype( (foo) ), void(&)()>::value, "" ); static_assert( std::is_same<decltype( (ti) ), int&>::value, "" ); 

Notice how this works for functions, and therefore in your case decltype(*&f) matches decltype( (f) ) , not decltype(f) .