Deprecated std::is_literal_type in C++17

问题

According to cppreference, the trait std::is_literal_type is deprecated in C++17. The question is why and what is the preferred replacement for the future to check whether a type is a literal type.

回答1:

As stated in P0174:

The is_literal type trait offers negligible value to generic code, as what is really needed is the ability to know that a specific construction would produce constant initialization. The core term of a literal type having at least one constexpr constructor is too weak to be used meaningfully.

Basically, what it's saying is that there's no code you can guard with is_literal_type_v and have that be sufficient to ensure that your code actually is constexpr. This isn't good enough:

template<typename T>
std::enable_if_t<std::is_literal_type_v<T>, void> SomeFunc()
{
  constexpr T t{};
}

There's no guarantee that this is legal. Even if you guard it with is_default_constructible<T> that doesn't mean that it's constexpr default constructible.

What you would need is an is_constexpr_constructible trait. Which does not as of yet exist.

However, the (already implemented) trait does no harm, and allows compile-time introspection for which core-language type-categories a given template parameter might satisfy. Until the Core Working Group retire the notion of a literal type, the corresponding library trait should be preserved.

The next step towards removal (after deprecation) would be to write a paper proposing to remove the term from the core language while deprecating/removing the type trait.

So the plan is to eventually get rid of the whole definition of "literal types", replacing it with something more fine-grained.

来源：https://stackoverflow.com/questions/40351816/deprecated-stdis-literal-type-in-c17