如何防止赋值运算符左侧出现临时性 [英] How to prevent a temporary on the left hand side of the assignment operator

问题描述

如果我为类型定义operator+，则采用通常的方式

If I define operator+ for a type, in the usual fashion

struct S {};

S operator+(S const &, S const &) { 
  return {}; 
}

S的用户可以编写类似

S s{};
s + s = S{}; // huh

据我所知，operator+返回类型为S的临时对象，然后将其分配给该对象.然后该对象在语句的末尾死亡，因为它没有名称，因此该语句实际上是无操作的.

From what I can tell, operator+ returns a temporary object of type S, which is then assigned to. The object then dies at the end of the statement, because there's no name for it, and so the statement is effectively a no-op.

我没有看到类似的代码用处，因此我想使它成为编译错误.有没有办法做到这一点?即使是警告也总比没有好.

I don't see any use for code like that, so I would like to make that a compile error. Is there a way to do that? Even a warning would be better than nothing.

推荐答案

刚刚找到了我需要的

Just found what I need here. Apparently I can make the assignment operator only bind to lvalues.

struct S {
    S& operator=(S const&) & // only binds to lvalues
    { 
      return *this; 
    }
}; 

现在我完全得到了我想要的错误.

Now I get exactly the error I want.

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