assigning to rvalue: why does this compile?

问题

In the following example:

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 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?
}

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+.

回答1:

(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;
}

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

来源：https://stackoverflow.com/questions/33784044/assigning-to-rvalue-why-does-this-compile