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Since you don't use i and j in the loops' body, you don't need nested loops at all.



for(sidx=0; sidx<72; sidx++)


As src[sidx] is far away from src[sidx+72] in memory, you're going to get...

S = 3*(x1+x2+..)+xk

and we want to find k.

Loop through the array again but subtract 3*xi from S
if i != k then (S-3*xi)%3 == xk
if i == k then (S-3*xi)%3 == -2*xk

So, in the first...

So, what you want to do is put the lowest bit of 0x42 into the highest of an
intermediary, unsigned byte. Put the top half of 0xF0 into the next four bits. And as the
five right-most bits make up...

sample_rate = sample_rate & 0xFFFFF0

Right shift by 4 is what you need to do here.

Alternatively, you could think of the string as a number in base 26 then convert to base 10 and back in the usual ways. Of course, this limits the allowed length of words, e.g., for uint32 the limit...

Don't you mean every multiple of 4? If so, using the not operator will help you get those zeroes. ;)

You could search mathworld for 'Square Root Algorithms.' There are a few methods described there for finding square roots.

I see; you have misread the code snippet I posted above.

Quzah, please elaborate.

The [i][j] syntax can be recovered by allocating an array of pointers, as well as the linear chunk, and setting each of these to point to each row:



char *outputWindowArray;
char...

"...Accept certain inalienable truths: prices will rise, politicians will philander..."

There's nothing new about the news.

Yes. The left most bit indicates the sign and when right shifting, the word may be filled with whatever the bit was--though, I think that's implementation defined.

Try changing int to unsigned int everywhere, and %d to %u.

Google 'strtod.'

It was a trick question and you've given the full solution. tsk tsk

Why not read the number as a string of digits?