Forcing a constant expression to be evaluated during compile-time?

Question

A few days ago I asked by which criteria the compiler decides whether or not, to compute a constexpr function during compile time.

When does a constexpr function get

Answer 1

Just to not leave it buried in comments:

#include <type_traits>

#define COMPILATION_EVAL(e) (std::integral_constant<decltype(e), e>::value)

constexpr int f(int i){return i;}

int main()
{
    int x = COMPILATION_EVAL(f(0));
}

EDIT1:

One caveat with this approach, constexpr functions can accept floating-point and be assigned to constexpr floating-point variables, but you cannot use a floating-point type as a non-type template parameter. Also, same limitations for other kinds of literals.

Your lambda would work for that, but I guess you would need a default-capture to get meaningful error message when non-constexpr stuff get passed to the function. That ending std::move is dispensable.

EDIT2:

Err, your lambda approach doesn't work for me, I just realized, how it can even work, the lambda is not a constexpr function. It should not be working for you too.

It seems there's really no way around it but initializing a constexpr variable in local scope.

EDIT3:

Oh, ok, my bad, the purpose of the lambda is just the evaluation. So it's working for that. Its result, instead, is which is unusable to follow another compilation time eval.

EDIT4:

On C++17, lambdas now can be used in constexpr contexts, so the limitation referred to in EDIT2/EDIT3 is removed! So the lambda solution is the correct one. See this comment for more information.