为什么不使用is_const类型特征将const引用视为const？ [英] Why isn't a const reference considered const using the is_const type trait?

问题描述

我对以下代码感到惊讶：

I was rather surprised that the following code:

#include <iostream>
#include <type_traits>

using namespace std;

int main(int argc, char* argv[]) {
    cout << boolalpha << is_const<const float&>::value << endl;

    return 0;
 }

打印 false 。删除引用可以正常工作：

Prints false. Removing the reference works correctly:

#include <iostream>
#include <type_traits>

using namespace std;

int main(int argc, char* argv[]) {
    cout << boolalpha << is_const<remove_reference<const float&>::type>::value << endl;

    return 0;
 }

打印出 true 。

两者都是使用G ++版本的 g ++ -std = c ++ 11 test.cpp 编译的： / p>

Both were compiled with g++ -std=c++11 test.cpp, using G++ version:

g++ (Ubuntu 5.3.0-1ubuntu1~14.04) 5.3.0 20151204

考虑一下，我可以理解这里有两种类型：引用类型和被引用的类型。引用的类型为 const ，因此第二种情况很有意义。对于第一种情况，如果引用的类型为 const 或始终为 true ，我希望它返回，因为引用AFAIK不能重新分配。

Thinking about it, I can understand that there are two types in play here: the reference type and the type that is referenced. The type that is referenced is const, so the second case makes sense. For the first case, I would expect it to either return if the referenced type is const or always true instead, because references AFAIK can't be "reassigned".

为什么它返回 false ？

推荐答案

结果正确。

引用绝不是 const ，因为它们不能通过 cv认证。您的说法是正确的，因为它们无法重新分配（因此在某种意义上是不变的），但这与 const 限定的含义不同。

References are never const, because they cannot be cv-qualified. You are correct in saying that they cannot be reassigned (and are, as such, immutable in a sense), but that is not the same as being const-qualified.

如果纯粹由于所指对象为 const <而得到 true / code>，那么在所有其他情况下，这将与 is_const 的含义完全不一致。例如考虑指针。 is_const< int const *> :: value 应该是什么？

If you got true purely because the referent is const, then that would be completely inconsistent with the meaning of is_const in all other cases. Consider, for example, pointers. What should is_const<int const*>::value be?

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