I’m looking at some code which should be trivial — but my math is failing me miserably here.
Here’s a condition that checks if a number if a power of 2 using the following:
if((num != 1) && (num & (num - 1))) { /* make num pow of 2 */ }
My question is, how does using a bitwise AND between num and num – 1 determine if a number is a power of 2?
Any power of 2 minus 1 is all ones: (2 N – 1 = 111….b)
Take 8 for example. 1000 & 0111 = 0000
So that expression tests if a number is NOT a power of 2.