关于C ++ 0x参考崩溃的问题 [英] Question about C++0x reference collapse

问题描述

我不知道为什么这些代码不能编译。我在Visual c + + 2010和gcc测试与-std = c + + 0x。任何人给一些建议？
谢谢！

I don't know why these code can't be compiled. I'v tested in Visual c++ 2010 and gcc with -std=c++0x. anyone give some suggestion? thanks!

template<typename T>
class Foo
{
public:
 void test(const T&){cout<<"const";}
 void test(      T&){cout<<"non const";}
};

int main()
{
 int a;
 Foo<int&> f;
}

编译错误：'void Foo :: test（T）'：成员函数已定义或已声明

compile error: 'void Foo::test(T)' : member function already defined or declared

但为什么可以编译？

template<typename T> void foo(const T&){cout<<"const"; }
template<typename T> void foo( T&){cout<<"non const"; }
int main()
 {
    int a; 
    foo<int&>(a);
 }

i'v read c ++ 0x article说： & == T& ，因此const T& & == const T&

i'v read c++0x article said: T& & ==T& , so const T& & == const T& ?

推荐答案

我读了c ++ 0x文章说： & == T& ，因此const T& & == const T& ？

i'v read c++0x article said: T& & ==T& , so const T& & == const T& ?

实际上，这没有什么意义。 IMHO，最好把它放在表格中：

Actually, that doesn't make a lot of sense. IMHO, it's better to put this into a table:

T       T&      const T      const T&
---------------------------------------
int     int&    const int    const int&
int&    int&    int&         int&
        (1)     (2)          (1+2)

1: Reference collapsing in action
2: const applies to the reference and is therefore ignored

如果T已经是引用（第二行） const T 适用于参考，而不适用于裁判。但是引用本质上是不变的，因为你不能在初始化后引用另一个对象，所以在这里忽略了一个 const 。你可以认为它是const崩溃。 ; - ）

If T is already a reference (2nd row) the const in const T applies to the reference and not to the referee. But a reference is inherently constant in the sense that you cannot make it refer to another object after initialization, so a const is just ignored here. You can think of it as "const collapsing". ;-)

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