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 :-)
