分配给右值:为什么要编译? [英] assigning to rvalue: why does this compile?

问题描述

在以下示例中:

class A {
  private: double content;

  public:
  A():content(0) {}

  A operator+(const A& other) {     
    content += other.content;
    return *this;
  }

 void operator=(const A& other) {
       content = other.content;
  }

};

A 是用于double的简单包装器，其 + 和 = 运算符已被重载.在以下用途中:

A is a simple wrapper for a double for which the + and = operators have been overloaded. In the following use:

 int main(int argc, char *argv[]) {
    A a, b, c;
    (a+b) = c ; // Why is this operation legal?
}

为什么(a + b)= c 编译?我想知道为什么此语句合法，因为(a + b))必须是 rvalue .我没有从 operator + 返回参考.

Why does (a+b) = c compile? I would like to know why this statement is legal, because the result of (a+b) must be an rvalue. I am not returning a reference from operator+.

推荐答案

(a + b)= c 与(a + b).operator =(c).赋值运算符中的右值引用没有特殊的规则，只是遵循通常的函数调用规则.如果要防止使用右值调用，可以添加一个ref-qualifier:

(a+b) = c is the same as (a+b).operator=(c). There is no special rule for rvalue references in assignment operators, it just follows usual function call rules. If you want to prevent calling with rvalues, you can add a ref-qualifier:

void operator= (const A& other) & {
//                              ^
     content = other.content;
}

这将仅允许在左值上调用该函数.

This will only allow the function to be called on lvalues.

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