如何将引用类型转换为值类型？

Question

I'm trying to move some code to templates using the new decltype keyword, but when used with dereferenced pointers, it produces reference type. SSCCE:

#include <iostream>

int main() {
    int a = 42;
    int *p = &a;
    std::cout << std::numeric_limits<decltype(a)>::max() << '\n';
    std::cout << std::numeric_limits<decltype(*p)>::max() << '\n';
}

The first numeric_limits works, but the second throws a value-initialization of reference type 'int&' compile error. How do I get a value type from a pointer to that type?

Answer 1

You can use std::remove_reference to make it a non-reference type:

std::numeric_limits<
    std::remove_reference<decltype(*p)>::type
>::max();

Live demo

or:

std::numeric_limits<
    std::remove_reference_t<decltype(*p)>
>::max();

for something slightly less verbose.

Answer 2

If you are going from a pointer to the pointed-to type, why bother dereferencing it at all? Just, well, remove the pointer:

std::cout << std::numeric_limits<std::remove_pointer_t<decltype(p)>>::max() << '\n';
// or std::remove_pointer<decltype(p)>::type pre-C++14

Answer 3

你想删除引用以及我猜的潜在的const ，所以你要使用

std::numeric_limits<std::decay_t<decltype(*p)>>::max()