操作符>智能指针的返回值 [英] the operator-> return value of smart pointers

问题描述

智能指针如shared_ptr可以像使用 * 和 - > 运算符的普通指针一样使用。
这些书说 - > 运算符返回shared_ptr存储的指针。所以你可以使用它来访问这个指针指向的对象。但我在这里感到困惑。看看下面的代码。

smart pointers like shared_ptr can be used like ordinary pointers with * and -> operator. The books say that -> operator returns the pointer that shared_ptr stores. So you can use it to access the object this pointer points to. But I am confused here. Look at the code below.

class A
{
public:
    A(int v = 20){val = v;}
    int val;
}
A* p1 = new A;
std::cout<<p1->val;  //This is common sense

boost::shared_ptr<A> p2(new A);
std::cout<<p2->val;  //This is right
//My question is that p2-> returns the pointers of the object, then maybe another 
//-> should be used?
//like (p2->)->val? 

推荐答案

好了，更像是一个特例。标准说

It's magic. Well, more like a special case. The standard says that

13.5.6类成员访问权限[over.ref]

1 operator-> 应为不带参数的非静态成员函数。它实现了使用 - > 的类成员访问语法。

1 operator-> shall be a non-static member function taking no parameters. It implements the class member access syntax that uses ->.

post fi x-expression - > templateopt id-expression

post fi x-expression - > pseudo-destructor-name

postfix-expression -> templateopt id-expression
postfix-expression -> pseudo-destructor-name

表达式 x-> m 解释为（x.operator->（）） - > ;如果 T 的类对象 x c $ c> T :: operator->（）
存在，并且如果操作符被重载解决机制（13.3）选择为最佳匹配函数。

An expression x->mis interpreted as (x.operator->())->m for a class object x of type T if T::operator->() exists and if the operator is selected as the best match function by the overload resolution mechanism (13.3).

也就是说，对重载的运算符的结果再次调用 operator-> 。如果那个被重载，它会递归地继续，直到一个原始指针是结果，并且调用内置 operator-> 。

That is, operator-> is called again on the result of the overloaded operator. And if that one is overloaded too, it goes on recursively until a raw pointer is the result and the built-in operator-> is called.

这并不意味着结果不能是任意类型 - 它可以是，但是你只能使用函数调用语法调用它：

That doesn't mean the result can't be any arbitrary type - it can be, but then you can only call it with the function call syntax:

struct X {
    int operator->() { return 42; }
};

int main()
{
    X x;
    x.operator->(); // this is OK
}

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