Subj : Re: possible overflow?
To   : comp.programming
From : Rob Somers
Date : Sat Aug 27 2005 09:00 am

> !0 = 1
> !1 = 0
> !2 = 0
> etc.
> 
> Another negation swaps the 0 and 1 again, so basically 0 remains 0,
> and any other value becomes 1.
> 
> 'mask' has a single bit set, each loop it's shifted one position.
> 
> If you AND 'input' with 'mask', the result is zero when the Nth bit
> of 'input' is clear, and it's non-zero when it's set.
> 
> So, !!(input & mask) equals 0 if the 'current' bit is clear,
> and 1 if it is set.
> 
> '0' is the ASCII value of the character 0.
> If you add one to that, you get the ASCII value of the character 1.
> 
> So, the end result is '0' if the bit is clear, and '1' if it is set.
> 
> 
> A direct and clear way to write this would be:
> 
>   putchar((input & mask) ? '1' : '0');
> 
> 
> But that might(*) be less efficient, so the double-negation trick
> is favoured by some programmers.
> However, the following might even be more efficient:
> 
>   putchar('1' - !(input & mask));
> 
> Try to work out for yourself how that one works.
> 
> 
> *) With any luck, a decent optimizing compiler will produce
>    equally fast code for the clear way, and it might even be faster.
>    For the record: gcc produces *the same* code for all three lines.
> 
> 
> SaSW, Willem

I feel enlightened now.  :)  Thanks for being so thorough in your
explanation.

Rob Somers

.