Question about over loading

[Udaya Bhaskar]

To differentiate between postfix and prefix opeators , when
overloaded, we need to pass a dummy parameter.

Why should we pass ONLY int to overload postfix ++ / -- operator?

Why not a FLAOT/ DOUBLE ?

or a bit field?

Thanks,
Uday.

 

[Victor Bazarov]

To differentiate between postfix and prefix opeators , when
overloaded, we need to pass a dummy parameter.

Why should we pass ONLY int to overload postfix ++ / -- operator?

Why not a FLAOT/ DOUBLE ?

or a bit field?

Why? Because the Standard says so, that's why. See 13.5.7.

V

 

[Rolf Magnus]

To differentiate between postfix and prefix opeators , when
overloaded, we need to pass a dummy parameter.

Why should we pass ONLY int to overload postfix ++ / -- operator?

Why not a FLAOT/ DOUBLE ?

or a bit field?

What advantage would that have?

 

[Mike Wahler]

To differentiate between postfix and prefix opeators , when
overloaded, we need to pass a dummy parameter.

Why should we pass ONLY int to overload postfix ++ / -- operator?

Why not a FLAOT/ DOUBLE ?

or a bit field?

Theoretically, the type doesn't really matter. We simply need
*something* in order to differentiate it. 'int' was
probably chosen because it's likely to be the most
efficient type to pass as an argument. Finally, we
must use 'int' because the language standard specifically
mandates it.

Why does it matter to you?

-Mike

 

[Michiel Salters]

Theoretically, the type doesn't really matter. We simply need
*something* in order to differentiate it. 'int' was
probably chosen because it's likely to be the most
efficient type to pass as an argument. Finally, we
must use 'int' because the language standard specifically
mandates it.

In fact, efficiency doesn't matter. No argument is actually passed
to the function, just as the =0 in a pure virtual funcion is not
actually an assignment.
int was probably chosen because the postfix ++ adds 1, and 1 is
an int. But your last point is basically what matters

HTH,
Michiel Salters

 

[Allan W]

(e-mail address removed) (Michiel Salters) wrote

In fact, efficiency doesn't matter. No argument is actually passed
to the function, just as the =0 in a pure virtual funcion is not
actually an assignment.
int was probably chosen because the postfix ++ adds 1, and 1 is
an int.

Sorry, that's not correct.

13.5.7 Increment and decrement
The user-defined function called operator++ implements the prefix
and postfix ++ operator. ...
If the function is a member function with one parameter (which
shall be of type int) or a non-member function with two parameters
(the second of which shall be of type int), it defines the postfix
increment operator ++ for objects of that type. When the postfix
increment is called as a result of using the ++ operator, the int
argument will have value zero.

There follows an example in which they call a.operator++(0) and
operator++(b,0) explicitly passing the value of the int.

// MY example, shorter than the one in the standard
#include <iostream>
using namespace std;
struct Foo {};
Foo &operator++(Foo &lhs, int rhs)
{ cout << "rhs is " << rhs << endl; return lhs; }
int main() {
Foo f;
f++; // rhs is 0
operator++(f,27); // rhs is 27
}

Using the second parameter in operator++ is unusual at best, and
obfuscated at worst -- but it's legal.

 

[chris]

To differentiate between postfix and prefix opeators , when
overloaded, we need to pass a dummy parameter.

Why should we pass ONLY int to overload postfix ++ / -- operator?

Why not a FLAOT/ DOUBLE ?

or a bit field?

The standard actually says that you shouldn't attempt to read from /
write to that parameter. To be honest, that int parameter is an ugly
hack to give people a way of overloading seperatly the postfix operator
(originally, you could only overload prefix, and the postfix operator
simply made a copy of the object and then did the prefix operator).

Now, just accept it as a slightly bizarre hack, and ignore the parameter
(it's best all around to simply not give it a name, and write
operator++(int) {..})

Chris