请说明以下功能

Question

I have come across a function definition starting as:

int operator*(vector &y)
{
  // body
}

After putting * just after operator and before opening brace of argument, what does this function mean?

Answer 1

This is an operator * overload. The syntax you should use is *(y) while y is of type vector .

It allows you a reference like implementation, something similar to pointer reference in C. Of course the actual meaning depends on the body. eg you can return a reference to an internal element in the vector.

Answer 2

这是*运算符的函数重载。

Answer 3

它的函数重载使解引用运算符*重载。

Answer 4

It is either a dereferencing operator or a multiplication operator override. It is dereferencing if it is in a namespace and multiplication if it is inside a class. Since it has a body and no class scope I will also assume that it is a dereferencing.

Answer 5

Actually its not a deferencing operator as in *ptr! Its actually an operator such as a multiplication operator. Here is a simple example

#include <iostream>
using namespace std;

struct Int{
 int val;
 Int(const int val = 0) : val(val){}
 int operator*(const Int& number)const{
    return val * number.val;
 }
};

int main(){
  Int n(4), m(5);
  cout << n * m << endl; //use the operator*() implicitly
  cout << (n.operator*(m)) << endl; //use the operator* explicitly
}

To define a de-ferenceing operator, its prototype would be operator*(). Look here for more information. Here is a live code to test.