二进制'*'：没有找到全局运算符，它接受类型“统计量”（或没有可接受的转换） [英] binary '*' : no global operator found which takes type 'statistician' (or there is no acceptable conversion)

问题描述

但是当我重载我的（*）时，我会重载我的运算符它真的只是一个类，它保存算术函数和一系列数组变量。乘法运算符我得到这个错误：

 二进制'*'：没有全局运算符找到类型'statistician'
 （或没有可接受的转换）
  

这种情况发生在我的代码尝试做：<$ c $ main.cpp中的c> s = 2 * u;

其中s和u是统计学类。

statistician =我的班级

（statistician.h）

  class statistician 
 {
 ...其他函数&变量... 
 
 const统计量statistician :: operator *（const statistician& other）const; 
 
 .....更多重载... 
 
}; 
  

任何帮助都是真棒的感谢!!

解决方案

声明命名空间作用域 operator * ，以便您也可以在不是统计学家类型的左侧

 统计运算符*（const统计数据和左数，const统计数据与右数）{
 // ... 
} 
  

不用说你应该删除类中的需要一个转换构造函数来接受 int 。

I am trying to overload my operators its really just a class that holds arithmetic functions and a sequence of array variables.

But when i am overloading my (*) multiplication operator i get this error:

     binary '*' : no global operator found which takes type 'statistician' 
(or there is no acceptable conversion)

This happens when my code tries to do: s = 2*u;in main.cpp

where s, and u are statistician classes.

statistician = my class

(statistician.h)

class statistician  
{
... other functions & variables...

const statistician statistician::operator*(const statistician &other) const;

..... more overloads...

};

Any help would be awesome thanks!!

解决方案

Declare a namespace scope operator*, so that you can also have a convertible operand on the left hand side that is not of type statistician.

statistician operator*(const statistician &left, const statistician &right) {
  // ...
}

Needless to say that you should remove the in-class one then, and you need a converting constructor to take the int.

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