⟩ How can I get random integers in a certain range?
The obvious way,
rand() % N /* POOR */
(which tries to return numbers from 0 to N-1) is poor, because the low-order bits of many random number generators are distressingly non-random.
A better method is something like
(int)((double)rand() / ((double)RAND_MAX + 1) * N)
If you'd rather not use floating point, another method is
rand() / (RAND_MAX / N + 1)
If you just need to do something with probability 1/N, you could use
if(rand() < (RAND_MAX+1u) / N)
All these methods obviously require knowing RAND_MAX (which ANSI #defines in <stdlib.h>), and assume that N is much less than RAND_MAX.
When N is close to RAND_MAX, and if the range of the random number generator is not a multiple of N (i.e. if (RAND_MAX+1) % N != 0), all of these methods break down: some outputs occur more often than others. (Using floating point does not help; the problem is that rand returns RAND_MAX+1 distinct values, which cannot always be evenly divvied up into N buckets.) If this is a problem, about the only thing you can do is to call rand multiple times, discarding certain values:
unsigned int x = (RAND_MAX + 1u) / N;
unsigned int y = x * N;
unsigned int r;
do {
r = rand();
} while(r >= y);
return r / x;