C++11: std::max(a,b) in static_assert()?

I noticed, that in [24.4.7] of the last C++-Std Doc N3291 max ist not constexpr:

template<class T> const T& max(const T& a, const T& b);

Therefore, it is not allowed to use it in a static_assert for example. Correct开发者_如何学Python?

static_assert( max(sizeof(int),sizeof(float)) > 4, "bummer" );

That is correct.

I imagine the reason is simply that std::max calls T::operator< for an arbitrary type T and for std::max to be constexpr, it would require T::operator< to be constexpr, which is unknown.

This is correct; std::min and std::max are not constexpr, not even in the latest draft of C++14 (N3690), so they cannot be used within constant expressions.

There is no good reason for this, only bad reasons. The most significant bad reason is that the C++ committee is composed of individuals who have a limited amount of time to work on standardization, and no-one has put in the work required to make these functions constexpr yet.

Note N3039, a change to the C++ standard adopted in 2010, that slightly extended the constexpr facility specifically so that function such as min and max could be made constexpr.

You can work around this by defining your own min and max functions:

template<typename T>
constexpr const T &c_min(const T &a, const T &b) {
  return b < a ? b : a;
}
template<typename T, class Compare>
constexpr const T &c_min(const T &a, const T &b, Compare comp) {
  return comp(b, a) ? b : a;
}
template<typename T>
constexpr const T &c_min_impl(const T *a, const T *b) {
  return a + 1 == b ? *a : c_min(*a, c_min_impl(a + 1, b));
}
template<typename T>
constexpr T c_min(std::initializer_list<T> t) {
  return c_min_impl(t.begin(), t.end());
}
// ... and so on

this works on c++ 11

template<const Sz a,const Sz b> constexpr Sz staticMax() {return a>b?a:b;}

use:

staticMax<a,b>()

and of course a and b must be constexpr