How to use unscoped enumerator as if its type is its underlying type

Question

This doesn't work as expected:

template<typename T>
struct PHI {
    enum : T { value = 11400714819323198485 >> (64 - sizeof(T) * 8) };
};

std::cout << PHI<unsigned long long>::value;

The output is 2135587861. What I expected is 11400714819323198485.(In VS2013)

I thought PHI<unsigned long long>::value would implicitly convert to type unsigned long long if needed. But it actually convert to unsigned. That means when I used it in some other place, it might convert to unsigned too. That's not what I want.

Answer 1

Let's get rid of the templates. Minimized repro:

#include <iostream>

enum bar : unsigned long long { baz = 11400714819323198485ULL };

void foo(int v) { std::cout << "int "<< v; }
void foo(unsigned v) { std::cout << "uint " << v; }
void foo(unsigned long long v) { std::cout << "ull " << v; }

int main() { foo(baz); }

This prints uint 2135587861.

Meanwhile,

#include <iostream>

enum bar : unsigned long long { baz = 11400714819323198485ULL };

void foo(unsigned long long v) { std::cout << "ull " << v; }

int main() { foo(baz); }

prints ull 11400714819323198485. So the value is preserved, and the conversion can be done. This looks like a bug in VC++'s overload resolution. This also reproduces in VS2015 CTP5.

Edit: reported as https://connect.microsoft.com/VisualStudio/feedback/details/1131433

The shortest workaround appears to be using foo(+baz); the unary + forces an integral promotion before overload resolution is performed.