Exchanging two integers

[Tomislav Novak]

Hi!

Is there a way I could exchange a and b (both a and b are integers) without using a temporary variable, so a = b and b = a? Some oneliner, or macro?
I think I saw once something with XOR, but I can't really remember what it was...

Thanks.

[Rodrigo]

Sure, using XOR

a ^= b ^= a ^= b;

or additions and subtractions:

a += b; a -= b = a - b;

I think there was a third way, but I can't remember it now.

[Tomislav Novak]

a ^= b ^= a ^= b;

or additions and subtractions:

a += b; a -= b = a - b;

Thanks! That's what I was looking for...

[junjieliang]

The third way that I know of is similar to the addition/subtraction method:

a = a * b;
b = a / b;
a = a / b;

Well then again this method is real stupid... you can get overflow and precision errors very easily. Just for fun I knowing it, but don't use it in real programs!

[Krzysztof Duleba]

Actually doing anything else than swaping the integers using a temporary variable is slower and shouldn't be used in code at all.
For stl classes, std::swap(T, T) is the best choice. For instance, swaping two lists is O(1).

[jakabjr]

besides, this is defined by ANSI C as undetermination:
a ^= b ^= a ^= b;
since u are trying to modify and access a variable in at the same point of execution.
it might work, but it's not portable.