This algorithm goes through as many iterations as there are set bits. So if we have a 32-bit word with only the high bit set, then it will only go once through the loop. In the worst case, it will pass once per bit. An integer n has log(n) bits, hence the worst case is … Read more
Unless you’re using an ancient compiler, it can already handle this level of conversion on its own. That is to say, a modern compiler can and will implement i % 2 using a bitwise AND instruction, provided it makes sense to do so on the target CPU (which, in fairness, it usually will). In other … Read more
First remember that machine words are of fixed size. Say 4, and that your input is: +—+—+—+—+ | a | b | c | d | +—+—+—+—+ Then pushing everything one position to the left gives: +—+—+—+—+ | b | c | d | X | +—+—+—+—+ Question what to put as X? with a … Read more
Any power of 2 minus 1 is all ones: (2 N – 1 = 111….b) 2 = 2^1. 2-1 = 1 (1b) 4 = 2^2. 4-1 = 3 (11b) 8 = 2^3. 8-1 = 7 (111b) Take 8 for example. 1000 & 0111 = 0000 So that expression tests if a number is NOT a … Read more
One thing to notice about bitmasks like that is that they are always one less than a power of two. The expression 1 << n is the easiest way to get the n-th power of two. You don’t want Zero to provide a bitmask of 00000001, you want it to provide zero. So you need … Read more
If you want the k-th bit of n, then do (n & ( 1 << k )) >> k Here we create a mask, apply the mask to n, and then right shift the masked value to get just the bit we want. We could write it out more fully as: int mask = 1 … Read more
Vexingly, if you were using double and long, there is BitConverter.DoubleToInt64Bits and BitConverter.Int64BitsToDouble. I have genuinely no idea why there aren’t Single / Int32 equivalents, as it forces you to create a pointless byte[] on the heap (it doesn’t even let you pass in a pre-existing buffer). If you are happy to use unsafe code, … Read more
Given that you know that every other bit is 0 in your application, you can do it like this: x = (x | (x >> 1)) & 0x33333333; x = (x | (x >> 2)) & 0x0f0f0f0f; x = (x | (x >> 4)) & 0x00ff00ff; x = (x | (x >> 8)) & 0x0000ffff; … Read more
the recursive solution to towers of Hanoi works so that if you want to move N disks from peg A to C, you first move N-1 from A to B, then you move the bottom one to C, and then you move again N-1 disks from B to C. In essence, hanoi(from, to, spare, N): … Read more
int lsr(int x, int n) { return (int)((unsigned int)x >> n); }