c++ - 基于 L 值与 R 值的重载

Question

我在一本 C++ 书中发现了以下内容:

Although we will not be doing it in this book, you can overload a function name (or operator) so that it behaves differently when used as an l-value and when it is used as an r-value. (Recall that an l-value means it can be used on the left-hand side of an assignment statement.) For example, if you want a function f to behave differently depending on whether it is used as an l-value or an r-value, you can do so as follows:

class SomeClass { 
public:
  int& f(); // will be used in any l-value invocation const 
  const int& f( ) const; // used in any r-value invocation ... 
};

我试过了，没用:

class Foo {
        public:
        int& id(int& a);
        const int& id(int& a) const;
};

int main() {
        int a;
        Foo f;
        f.id(a) = 2;
        a = f.id(a);
        cout << f.id(a) << endl;
}

int& Foo :: id(int& a) {
        cout << "Bar\n";
        return a;
}

const int& Foo :: id(int& a) const {
        cout << "No bar !\n";
        return a;
}

我是不是理解错了？

Answer 1

要么书中的示例完全错误，要么您从书中复制了错误的示例。

class SomeClass { 
public:
  int& f(); // will be used in any l-value invocation const 
  const int& f( ) const; // used in any r-value invocation ... 
};

使用这段代码，当您调用 s.f() 时，其中 s 是 SomeClass 类型的对象，第一个版本将在s 是非const，当s 是const 时将调用第二个版本。值类别与它无关。

Ref 资格看起来像这样:

#include <iostream>
class SomeClass {
public:
  int f() & { std::cout << "lvalue\n"; }
  int f() && { std::cout << "rvalue\n"; }
};
int main() {
    SomeClass s; s.f(); // prints "lvalue"
    SomeClass{}.f(); // prints "rvalue"
}