Now this may be silly..

 But yes, i didn't think its safe to swap two numbers numbers represented in fixed bits without using a temporary variable

We have this famous trick from college days,

$a = a + b$

$b = a - b$ # a + b - b = a

$a = a - b$ # a + b -a = b

Which i practically never needed to use, up until in a recent hobby project where the number of registers available are only two.

well my college days happened some 20 years back, so maybe i understood it at that time but not any more... use it or lose it you see.. 

When i wanted to use this trick now, what really concerned me was the overflow, and turns out the overflow would not matter. i just try to explain in simple terms as below.. 

 

For an $n$ bit width field, the largest unsigned number that can be represented is $2^n - 1$.

So for an overflow to occur on adding numbers $a$ and $b$, we would need $a$ as:

$a = 2^{n-1} + x$

and $b$ such that $x + b = 2^{n-1} + y$ where $y \ge 0$ 

From above, we can re-express $b$ as $b = 2^{n-1} + y - x$ Lets remember this as $eq(1)$ 

When we do

$a := a + b$, we have

$a := 2^{n-1} + x + b = 2^{n-1} + 2^{n-1} + y = 2^n + y$

But because we only have $n$ bits, the value stored at a would be:

$a = y$

And then we set out to do 

$b : = a - b = y - b$, so how would we do '$a - b$'?

We would first take 2's compliment of $b$ and then add it to $a$

2's compliment of $b$ is $2^n - b$

Let's now substiture $b$ from $eq(1)$, then we have,

$b := a - b = y - b = y + 2^n - b = y + 2^n - (2^{n-1} + y - x)$

Thus we get

$b := \cancel{y} + 2^n - 2^{n-1} - \cancel{y} + x = 2^{n-1} + x$

Now this last expression is exactly the same as the value originally at $a$.

Similary, we can convince ourselves for the step $a = a - b$ that follows 

But does it mean i understand the elegance of the modular arithmetic involved here?? No way :-)