为什么不能确定const T&是const? [英] Why is `const T&` not sure to be const?

问题描述

template<typename T>
void f(T a, const T& b)
{
    ++a; // ok
    ++b; // also ok!
}

template<typename T>
void g(T n)
{
    f<T>(n, n);
}

int main()
{
    int n{};
    g<int&>(n);
}

请注意:b属于const T&，++b可以！

为什么const T&不确定是常量?

Why is const T& not sure to be const?

推荐答案

欢迎使用const和引用崩溃.当您拥有const T&时，引用将应用于T，const也是如此.您像这样呼叫g

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

，因此您已指定T为int&.当我们将引用应用于左值引用时，两个引用会折叠为一个引用，因此int& &变为int&.然后我们从 [dcl.ref]/1 中了解规则，其中指出，如果将const应用于引用，则将其丢弃，因此int& const变为int&(请注意，您实际上不能声明int& const，它必须来自typedef或模板).这意味着

so you have specified that T is a int&. When we apply a reference to an lvalue reference, the two references collapse to a single one, so int& & becomes just int&. Then we get to the rule from [dcl.ref]/1, which states that if you apply const to a reference it is discarded, so int& const just becomes int& (note that you can't actually declare int& const, it has to come from a typedef or template). That means for

g<int&>(n);

您实际上是在打电话

void f(int& a, int& b)

您实际上并没有在修改常量.

and you are not actually modifying a constant.

您曾称g为

g<int>(n);
// or just
g(n);

然后T将是int，并且f将被标记为

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

由于T不再是引用，因此将const和&应用于该引用，并且由于尝试修改常量变量，您将收到一个编译器错误.

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

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