重载算术运算符 [英] Overloading Arithmetic Operators

问题描述

赋值运算符可以声明为

T& operator =（const t&）;

T& operator= (const t&);

，但算术运算符不能这样定义。它必须是朋友的功能。我不明白为什么？

in a class, but the arithmetic operators cannot be defined that way. It has to be friend function. I don't understand why? Can you please explain ?

推荐答案

算术运算符不应该是朋友

It is not mandatory that arithmetic operators should be friend

你可以这样定义：

MyClass MyClass::operator + (const MyClass& t) const
{
  MyClass ret(*this);
  ret += t;
  return ret;
}

a + b 真的是语法糖，编译器会将其扩展为 a.operator +（b）。如果所有对象都是MyClass实例，上一个示例将会工作，但是如果你必须使用其他类型的对象，即 1 + a

The a + b is really a syntax sugar, the compiler will expand it to a.operator+(b). The previous sample will work if all your objects are MyClass instances, but will not work if you have to operate with others types, ie 1 + a, will not work, this can be solved by using friends.

MyClass operator + (int i, const MyClass& t)
{
  MyClass ret(i);
  ret += t;
  return ret;
}

这必须在+操作符的左边不是一个类，或者它是一个类，但是你不能将operator +添加到它的定义。

This has to be done when the left hand side of the + operator is not a class, or it is a class but you can't add operator + to its definition.

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