“必须有类或枚举类型的参数”实际上是什么意思

问题描述：

我有一个头文件和一个.cpp文件。我需要写我的.h文件的功能，但我得到一个错误，我可以完全完成一个骨架的.cpp文件。

I have a header file and a .cpp file. I an needing to write functions for my .h file but i get an error before i can fully complete a skeleton .cpp file.

Money.h

#ifndef MONEY_H
#define MONEY_H

#include <iostream>
#include <iomanip>

using namespace std;

class Money
{
public:
   Money(int dollars, int cents);
   Money operator+(const Money& b) const;
   Money operator-(const Money& b) const;
   Money operator*(double m) const;
   Money operator/(double d) const;

   void print() const;

private:
   int dollars;
   int cents;
};

#endif

Money.cpp

#include "Money.h"

Money::Money(int dollars, int cents){

}
Money operator+(const Money& b) {

}
Money operator-(const Money& b) {

}
Money operator*(double m) {

}
Money operator/(double d) {

}

void print(){

}

错误是乘法和除法运算符：

The errors are with the multiply and divide operators:

Money.cpp：12：25：error：'Money operator *（double）'must have a
argument of class或枚举类型

Money.cpp:12:25: error: 'Money operator*(double)' must have an argument of class or enumerated type

Money.cpp：15：25：error：'Money operator /（double）'必须有类的
参数或枚举类型

Money.cpp:15:25: error: 'Money operator/(double)' must have an argument of class or enumerated type

答

您没有使用范围解析操作符告诉编译器您正在定义成员函数。它被解释为一个全局运算符重载，它接受两个参数，其中一个必须是类或枚举类型。这基本上意味着您的一个参数必须是用户定义的类型（不是原始类型）或枚举类型，它通过枚举定义。

You're not using the scope resolution operator to tell the compiler that you are defining a member function. It is instead interpreted as a global operator overload, which takes two arguments, one of which must be of class or enumerated type. This basically means that one of your arguments must either be a user-defined type (type that is not a primitive type) or an enumerated type which is defined through an enum.

在您的原始代码 Money 只是返回类型;它不会告诉编译器您正在从该类定义成员函数。

In your original code Money is just the return type; it doesn't tell the compiler that you are defining member function from that class.

这里是您的一行的修复：

Here is a fix for one of your lines:

Money Money::operator+(const Money& b)                                         /*
      ^^^^^^^                                                                  */
{
     // ...
}

此外，您的原型和定义也必须匹配cv资格。您的定义缺少 const 限定符...

Moreover, your prototypes and definitions must also match in cv-qualification. Your definitions were missing the const qualifier...

Money Money::operator+(const Money& b) const                                   /*
                                       ^^^^^                                   */
{
     // ...
}

更新：

Update:

发现您的 Money :: operator * 和 Money :: operator / 的定义与其原型不匹配。两个原型都采用 double ，而定义采用 Money const& 。

I also found that your definition for Money::operator* and Money::operator/ do not match their prototypes. The prototypes for both take a double while the definitions take Money const&. You will need to change one to match the other.

// inside Money class

Money operator*(Money const&) const;
Money operator/(Money const&) const;

“必须有类或枚举类型的参数”实际上是什么意思

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