^ operator?

[Dark Nexus]

i can remember reading about this in one of my books, but can't remember for the life of me what the heck it actually does. i saw it used to swap variables without using a temporary variable

 

int a = 1, b = 0;

a ^= b ^= a ^= b;

 

this sets a to 0 and b to 1

 

obviously it is overloaded so the original would be a = a ^ b;

 

anyone know?

[»SOS]

Well, it's a funky one A very useful one, too, if you like doing weird and funky stuff (and math/crypto).

 

It's the one and only exclusive or.

 

source ^ modifier = result

1001 ^ 1001 = 0000

1001 ^ 0110 = 1111

1001 ^ 0101 = 1100

 

If a source or modifier bit is one (but not both), the resulting bit is one.

[Dark Nexus]

well out of the context of 0's and 1's, i use it to swap out characters in a string, it works, and i just don't see how it works. maybe explain in being used to swap out characters in a character array?

 

thanks for the help

[»SOS]

You mean how you can do the thing in the first post?

Well, it all comes down to the ones and zeroes, really

 

byte a = 133, b = 44;

a ^= b ^= a ^= b;

 

What this does is as follows:

 

Note: a + b does not mean a sum here. It means a sort of a mix.

 

First, a is a and b is b.

The first op (a ^= causes a to become a mix of a and b (a is now a +

The second op (b ^= a) causes b to become a! What it does is this: it takes a (which is a + and then removes itself ( from that value. So a + b becomes just a.

And the last op (a ^= removes the a part (which is now in from a + b (which is in a) and a becomes only b.

[Dark Nexus]

ah i see, okies thank you very much