二进制'*'：找不到全局运算符，它接受类型'统计者'（或者没有可接受的转换）

Question

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!!

Answer 1

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 .

Answer 2

This is exactly why binary operators like * or + should be non-member.

If you did s = u * 2 , it would have worked, assuming that you have a non-explicit constructor for statistician that takes a single int argument. However, 2 * u does not work, because 2 is not a statistician , and int is not a class with a member operator* .

For this to work right, you should define a non-member operator* and make it a friend of statistician :

statistician operator*(const statistician &left, const statistician &right);

You also need to either define other versions of operator* that take integers (or whatever other types you wish to be able to "multiply") or define non-explicit constructors for statistician to enable implicit conversion.