is_const< const int&> :: value is false - 为什么？

问题描述：

为什么这个静态断言触发？

Why does this static assertion fire?

static_assert(std::is_const<const int&>::value, "Pain");

要获得一个句法（为什么实现这样做）和语义推理为什么他们会设计这种类型的trait的接口来做到这一点）。

Would be awesome to get both a syntactic (why the implementation would do this) and a semantic reasoning (why they would have designed this type trait's interface to do this).

我知道可以抛出一个 std :: remove_reference 调用获得预期结果，但我不知道为什么这是必要的。

I am aware it is possible to throw in a std::remove_reference call to get the expected outcome, but I'm not sure why that's necessary.

答

const int& 是对 const int 的引用。所以引用本身不是 const 。

const int& is a reference to const int. So the reference itself isn't const.

这有点混乱，因此我要提出一个类比与 const int * 。它是指向 const int 的指针。但你可以修改它

It's slightly confusing, so I'm going to present an analogy with const int*. It's the pointer to const int. But you can modify it

const int a = 5, b = 7;
const int* ptr = &a;
ptr = &b; // pointer is modified

因此指针不是 const 。（const指针将 int * const ）

so the pointer isn't const. (the const pointer would be int* const instead)

is_const< const int&> :: value is false - 为什么？

相关推荐