Why are type_traits implemented with specialized template structs instead of constexpr?

Question

Is there any reason why the standard specifies them as template structs instead of simple boolean constexpr?

In an additional question that

Answer 1

One reason is that constexpr functions can't provide a nested type member, which is useful in some meta-programming situations.

To make it clear, I'm not talking only of transformation traits (like make_unsigned) that produce types and obviously can't be made constexpr functions. All type traits provide such a nested type member, even unary type traits and binary type traits. For example is_void::type is false_type.

Of course, this could be worked around with std::integral_constant()>, but it wouldn't be as practical.

In any case, if you really want function-like syntax, that is already possible. You can just use the traits constructor and take advantage of the constexpr implicit conversion in integral_constant:

static_assert(std::is_void(), "void is void; who would have thunk?");

For transformation traits you can use a template alias to obtain something close to that syntax:

template 
using enable_if = typename std::enable_if::type;
// usage:
// template  enable_if(), int> f();
//
// make_unsigned x;