什么导致 C++ 编译器错误:必须有类或枚举类型的参数? [英] What causes C++ compiler error: must have argument of class or enumerated type?

问题描述

函数声明:

template <typename T>
Point<T>* operator +(Point<T> const * const point, Vector<T> const * const vector);

我已经有一段时间没有使用 C++了，所以也许我正在做一些非常愚蠢的事情.让我知道.

It's been a while since I've used C++ so maybe I'm doing something really stupid. Let me know.

另外，不，我没有使用命名空间标准.

Also, no, I am not using namespace std.

推荐答案

您在语言级别上做错的是重载指针运算符.重载运算符的至少一个参数必须是用户定义的类型，或者是对一个的引用.

What you're doing wrong here on the language level is overloading operators for pointers. At least one argument of an overloaded operator must be of a user-defined type, or a reference to one.

但你在另一个层面上也做错了.您正在返回一个指针，这意味着您可能需要在运算符中动态分配一些存储空间.那么，谁拥有该存储空间?谁来发布?

But you're also doing this wrong on another level. You're returning a pointer, which means you will probably need to allocate some storage dynamically in the operator. Well, who owns that storage? Who will release it?

您应该只获取引用并按值返回，例如:

You should just take references and return by value, something like:

template <typename T>
Point<T> operator +(Point<T> const& point, Vector<T> const& vector) {
    return Point<T>(point.x + vector.x, point.y + vector.y);
}

这篇关于什么导致 C++ 编译器错误:必须有类或枚举类型的参数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！