此函数如何成为重载运算符？

Question

this is a program to convert an object having a string containing a date and another object having 3 ints: date, month and year.

class date
{
private:
char dt[9];

public: 
//constructors and functions....
};

class dmy
{
private: 
int day,mth,yr;

public:
//constructors..

  operator date()   // This is what my question is about
  {
   char temp[3],str[9]
   itoa(day,str,10)
   strcat(str,"/");
  itoa(mth,temp,10);
  strcat(str,temp)
  strcat(str,"/");
  itoa(yr,temp,10);
  strcat(str,temp);

  return (date(str));
  }

};

int main
{
 date d1;
 dmy d2(17,11,94);

 d1=d2;

 //display d1's and d2's data
 return 0; 
}

My confusion lies in the operator date() part. how can date be an operator? shouldn't the = operator be overloaded to carry out this conversion?

Answer 1

A function of this form of

operator T() { /*return an object of type T*/ }

is called user-defined conversion function, which is invoked implicitly when an object of type T is required in any expression.

In your case, the user-defined conversion function is defined to implicity convert an object of type dmy into an object of type date .

void f(date dt);

dmy dmyObj; 

date dateObj = dmyObj; //ok. conversion function is invoked implicitly
f(dmyObj); //ok. conversion function is invoked implicitly

Answer 2

operator date() 

是一个转换函数 ，无论您的代码中什么地方需要date类型，而要使用dmy类型，则将隐式调用此运算符以创建date类型并使用它。