decltype appears to drop const qualifiers for overloaded operators by Mooing Duck

Thank you for contacting Microsoft. We now correctly give an error on your example case as the type of "v" is "X const &" while the argument of operator> is X& and type deduction for operator< fails when T=X and U=int is substituted in.

Tanveer Gani
Visual C++ Team.

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Here's a snippet that correctly fails to compile in VS11 beta:

struct X{};

bool f(X& a, int const& b){ return true; }

template<class T, class U>
auto g(T const& v, U const& u) -> decltype(f(v,u)){ return true; }

int main(void) {
X x;
g(x, 6);
}

One can even change the original code to use the free function calling form in the trailing return type and, again, it correctly fails to compile:

struct X{};

bool operator>(X& a, int const& b){ return true; }

template<class T, class U>
auto operator<(T const& v, U const& u) -> decltype(operator>(v,u)){ return true; }

int main(void) {
X x;
x < 6;
}

Somehow, overload resolution inside a trailing return type's decltype discards cv qualifiers on all arguments. Here's another snippet that shows the strange behaviour:

struct X{};

bool operator>(X const& a, int& b){ return true; }

template<class T, class U>
auto operator<(T volatile& v, U volatile& u) -> decltype(v > u){ return true; }

int main(void) {
X x; int i = 6;
x < i;
}