Programming

[Nanti-SARRMM]

Fred, my TA's are crazy. If you have two variables while programming in C, X and Y, and you want to exchange their values but cannot use a temporary variable, how would you do it?

I would say something like
int x, y;
{
x=x+y; y=x-y; x=x-y;
}
But the TA's say that would not work with latch numbers, but that somehow doing this works:
int x, y;
{
y=x^y;
x=x^y;
y=x^y;
}
This assumes that x and y are initialized and all that... so am I right or how would it be done?

[Fredjikrang]

Sorry, but I don't know squat about programming. I'm learning a little bit, but I'm not into any of that stuff. Yellow would know, or maybe Bismark.

[Giovanni Schwartz]

Off topic!

[Fredjikrang]

Sorry about that Giovanni! Didn't even realize that it was in the nomination thread!

[Nanti-SARRMM]

is that where i put it? I thought I put it in Random Chatter. Oh well.

[bismark]

define latch numbers?

the TA works like so (d= decimal, b= binary):

x = 3d = 011b
y = 4d = 100b

y = x ^ y = 111b
x = x ^ y = 100b = 4d
y = x ^ y = 011b = 3d

[Nanti-SARRMM]

define latch numbers?

I assume the latch number is the biggest number it can get before it overflows. Thanks for the help though.

[bismark]

ah.. never heard such a term.

so lets say for a 4 bit number using 2s compliment...

x = 7 = 0111
y = -8 = 1000
x = x + y = 0111 + 1000 = 1111 = -1
y = x - y = 1111 - 1000 = 1111 + 1000 = 0111 = 7
x = x - y = 1111 - 0111 = 1111 + 1001 = 1000 = -8

so that worked.....

x = -1 = 1111
y = -8 = 1000
x = x + y = 1111 + 1000 = 0111 = 7
y = x - y = 0111 - 1000 = 0111 + 1000 = 1111 = -1
x = x - y = 0111 - 1111 = 0111 + 0001 = 1000 = -8

hmm... did they give you the case where it doesn't work?

[361]

Both will work in C/C++ provided you are working with ints or int pointers. (with some caveats I'll point out later)

If you try with other variable types you might get implied type-casting or other un-expected behavior which could screw with your variables values before and after the operation. (like floats / chars / etc...) This is compiler dependent... so what may work on GCC won't behave the same on Intel's compiler...

The XOR method is, however, superior to addition/subtraction in that it can account for overflow...

i.e. If you try this with the following numbers:

x = 4294967295;
y = 1;

x = x + y = 4294967296 (which requires 33-bits to properly represent and will overflow) = ??? (undefined behavior here...)
So if the computer just chops off the 33nd bit... you get 0... so...
x = 0
y = 0 - 1 = -1
x = 0 - (-1) = 1

Using the XOR method we get
x = 11111111111111111111111111111111b
y = 00000000000000000000000000000001b

y = x ^ y = 11111111111111111111111111111110b
x = x ^ y = 00000000000000000000000000000001b
y = x ^ y = 11111111111111111111111111111111b

So although the addition and subtraction will work in most cases... XOR works 100%

I think I may have screwed up that example... but whatever... you get the idea...

[361]

It's also important to note that compilers are smart enough to recognize a temp variable swap pattern and perform optimizations for it...

So by specifying a two variable swap you may actually be making your code less efficient...

[bismark]

ah, 361, my bad, i didn't take into account unsigned integers.

so with a 4bit unsigned integers,

x = 1111 = 16
y = 1110 = 15

x = x + y = 1111 + 1110 = 0001 = 1
y = x - y = 0001 - 1110 = 0001 + 0010 = 3
x = x - y = 0001 - 0010 = 0001 + 1110 = 16

so yeah, ugly. assuming signed, 2s complement signed integers, even overflow shouldn't be an issue though as seen in my above example and here:

x = 7 = 0111
y = 1 = 0001

x = x + y = 0111 + 0001 = 1000 = -8
y = x - y = 1000 - 0001 = 1000 + 1111 = 0111 = 7
x = x - y = 1000 - 0111 = 1000 + 1001 = 0001 = 1

of course not all architectures use 2s compliment nor can we assume signed integers, so yes, the XOR is correct.

in C, there should be no difference between using a char and an integer with either of these functions. Floats are a totally different story, and XORing them would definitely not work. I don't feel like figuring out if the adding/subtracting method would.

[361]

The problem with overflows is the undefined behavior

You just can't guarantee it will behave the same...

[bismark]

yes, i have a terrible habit of picking terrible examples. and i stand corrected. again.

x = 0111 = 7
y = 0111 = 7

x = x + y = 0111 + 0111 = 1000 = -1
y = x - y = 1000 - 0111 = 1000 + 1001 = 0001 = 1
x = x - y = 1000 - 0001 = 1000 + 1111 = 0111 = 7

[Nanti-SARRMM]

Oh.. I get it then. Thanks all.

[Giovanni Schwartz]

I can't help. I don't speak nerd.


; )

[Fredjikrang]

Don't worry. If you want to be an engineer, you'll learn it soon enough. ;D

[361]

I can't help. I don't speak nerd.


; )

It's one of those languages... Minutes to learn... Lifetimes (multiple...) to master...

To begin a strong background in Leetspeak is recommended..

[yellow m&m]

I'm going to be learning this soon enough...

[Giovanni Schwartz]

101! Now I feel 31337.


(that 101 was lol. Intentionally written to be... Mind blank, can't think of word... Means attempting to be skilled, when your really not.)

Also, if you enjoy physics, you might like flashphysicsgames.com They have some fun ones. And some lame ones.
But mostly fun.