操作符重载 [英] Operator overloading

问题描述

为什么重载operator =强制为成员函数（$ 13.5.3），而不是复合赋值运算符。 operator + =（$ 13.5.2）？ c> $

作为成员，如果用户没有定义一个，它总是由编译器提供。我相信这只是为了简单和避免意想不到的歧义，它要求 operator = 不能被定义为自由函数。

转换运算符处理从用户定义类型到内置类型的转换。

Why is overloaded operator= mandated to be a member function ($13.5.3), but not a compound assignment operator e.g. operator+= ($13.5.2)? Am I overlooking something here?

解决方案

A copy assignment operator=, as a member, is always provided by the compiler if the user doesn't define one. I believe that it was only for simplicity and to avoid unexpected ambiguities that it was made a requirement that operator= can't be defined as a free function.

Conversion operators take care of the case when you want to assign from a user-defined type to a built-in type.

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