11

please consider following code

#include <iostream>
using namespace std;
class Digit
{

private:
    int m_digit;
public:
    Digit(int ndigit=0){
     m_digit=ndigit;
                        }
    Digit& operator++();//prefix
    Digit& operator--();   //prefix
        Digit operator++(int);
        Digit operator--(int);
        int get() const { return m_digit;}
};
Digit& Digit::operator++(){

   ++m_digit;
   return *this;
}
Digit& Digit::operator--(){
 --m_digit;
 return *this;

}
Digit Digit::operator++(int){
Digit cresult(m_digit);
++(*this);
return cresult;


}
    Digit Digit::operator--(int){
Digit cresult(m_digit);
--(*this);
return cresult;


}
    int main(){

     Digit cDigit(5);
      ++cDigit;
        cDigit++;
         cout<<cDigit.get()<<endl;
         cout<<cDigit.get()<<endl;





     return 0;
    }

here is implemented two version of postfix and prefix operators,i have read that difference is made by introduce another so called dummy argument,but i have question if we see declaration of these

Digit& operator++();//prefix
             Digit& operator--();   //prefix
        Digit operator++(int);
        Digit operator--(int);

they are differed by & mark,so why it is necessary dummy argument?and also in both case for example ++ operator is written before the argument and does not it means that they are same?

iammilind
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4 Answers4

15

The pre- and post-increment are two distinct operators, and require separate overloads.

C++ doesn't allow overloading solely on return type, so having different return types as in your example wouldn't be sufficient to disambiguate the two methods.

The dummy argument is the mechanism that the designer of C++ chose for the disambiguation.

NPE
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  • I imagine there is a notation in the arm or somewhere as to why this was the method chosen to disambiguation. The answer is probably as simple as "there was no other way." – John Dibling Oct 12 '11 at 13:16
8

In Pre-Increment/Decrement and Post-Increment/decrement, the difference is based on only dummy parameter in overloaded function

operator++()         => Prefix Increment
operator--()         => Prefix Decrement

operator++(int)      => Postfix Increment
operator--(int)      => Postfix Decrement

return type may be same. you can also refer: http://www.tutorialspoint.com/cplusplus/increment_decrement_operators_overloading.htm

Eswaran Pandi
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4

The operator, just as any function, is identified by the signature. The return type and modifiers before the function/operator name is not included in this. Your operators names are here

operator++()
operator++(int)

This is the only way the compiler distinguishes between the two. As for the Digit& and Digit return values; They are needed because of the way ++x, and x++ are supposed to operate.

Captain Giraffe
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2

In C++ functions/methods can't be overloaded by return type, only by parameter list. Ignoring the fact that the prefix and postfix operators are operators, imagine if they were just simple other functions, how would the compiler work out which to use based on the return type? E.g.

int x = 2;

const int DoIt()
{
    return 1;
}

int& DoIt()
{
    return x;
}

int y = DoIt();

Since operator overloads are really just functions at heart, there's no way for the compiler to differentiate between them by return type.

See http://www.parashift.com/c++-faq-lite/operator-overloading.html#faq-13.14

blaaaaaaah
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