C ++运算符重载将指针类型作为参数吗? [英] C++ operator overloading takes pointer type as parameter?

问题描述

我是C ++的新手，试图弄清楚指针和引用之间的区别.我刚刚阅读了

I'm new to C++ and trying to figure out the differences between pointer and reference. I've just read this short summary.

在文章中，作者提到day *operator++ (day *d);不会编译(注意:day是一个枚举类型)，并指出此重载运算符函数的参数必须为T，T&或T const& ;，其中T是类或枚举类型.

In the article, the author mentioned that day *operator++ (day *d); won't compile (note: day is an enum type) and argued that the parameter for this overloaded operator function must be type T, T&, or T const&, where T is a class or enum type.

我假设指针是内置类型，而不是类或枚举，因此它不能用于重载运算符，并且对于所有内置类型(如int和double)来说，运算符重载都是不可能的.

I assume that pointer is a built-in type rather than a class or enum so it can't be used to overload operators and that operator overloading is not possible for all built-in types such as int and double.

例如，int i = 1; ++i;永远不会通过重载类型为int的++运算符来使i为3.

For example, int i = 1; ++i; would never result in i being 3 by overloading the ++ operator for the type int.

我正确吗?请帮助我更好地理解这个问题.

Am I correct? Please help me understand this problem better.

推荐答案

运算符重载的第一个规则是:
您不能对内置数据类型重载运算符，只能对自定义数据类型重载，因此在这方面您是正确的.

First rule in Operator overloading is:
You cannot overload operators for built-in data types, You can only for your custom data types, So you are correct in that regard.

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