i++ and ++i when the ++ operator is overriden

[Binary]

Hi,

In this board's FAQ, the operator override section says the i++ and ++i
has some difference, the i++ will make a copy of this pointer.
I can't know the reason, why?

Thanks in advance.
ABAI

 

[Kai-Uwe Bux]

Hi,

In this board's FAQ, the operator override section says the i++ and ++i
has some difference, the i++ will make a copy of this pointer.
I can't know the reason, why?

i++ is supposed to return the value of i *before* incrementing. So, if you
implement postincrement for a class, the typical implementation looks
somewhat like this:

class SomeIterator {

// ...

SomeIterator operator++ ( int ) {
// save the old value
SomeIterator return_value ( *this );
// increment
++ *this;
// return the old value
return( return_value );
}

};


Best

Kai-Uwe Bux

 

[Binary]

Perfect! Thank you!

ABAI

i++ is supposed to return the value of i *before* incrementing. So, if you
implement postincrement for a class, the typical implementation looks
somewhat like this:

class SomeIterator {

// ...

SomeIterator operator++ ( int ) {
// save the old value
SomeIterator return_value ( *this );
// increment
++ *this;
// return the old value
return( return_value );
}

};


Best

Kai-Uwe Bux

 

[Michiel.Salters]

Hi,

In this board's FAQ, the operator override section says the i++ and ++i
has some difference, the i++ will make a copy of this pointer.

/should/ make a copy. Each built-in operator++(int) makes a copy, and
your
versions should behave the same. The same applies to operator==; your
versions should return a bool. Else, people who use it would be
surprised.

I.e. it's not a technical requirement but a usability reason.

HTH,
Michiel Salters