Keyword typeof in c++11

Question

I know:

• typeof is a gcc extension and is not part of the C++ standard.

Questions:

1. Is the word typeof deprecated in C++11? in other words, is it allowed to still use it as a gcc extension when using C++11?
2. Is it correct to say that replacing every typeof with decltype yields the same behaviour of the code?
3. Assume I have template<typename T> class wrapper. What is the best way to declare wrapper_some_field such that it is equivalent to: Wrapper<typeof(some_field)> wrapper_some_field

Answer 1

Is the word typeof deprecated in C++11? in other words, is it allowed to still use it as a gcc extension when using C++11?

It's not deprecated. It never existed as a keyword. gcc suggests that if you compile with -std=c++** that you instead use __typeof__.

Is it correct to say that replacing every typeof with decltype yields the same behaviour of the code?

No. For example, given:

int& foo();

decltype(foo()) is int& but __typeof__(foo()) is int.

Assume I have template<typename T> class wrapper. [...]

You could write wrapper<std::remove_reference_t<decltype(some_field)>> wrap{some_field}, but it'd be cleaner to write a construction function template:

template <class T> wrapper<T> make_wrapper(T const& val) { return {val}; }
auto wrap = make_wrapper(some_field);

Or, with forwarding:

template <class T>
wrapper<std::decay_t<T>> make_wrapper(T&& val) {
    return {std::forward<T>(val)};
}

Although in C++17 you wouldn't do this at all and would just use class template argument deduction:

template <class T> struct wrapper { T t; };
template <class T> wrapper(T) -> wrapper<T>;
auto wrap = wrapper{42}; // wrap is a wrapper<int>

And in C++20, you won't even need the deduction guide.