1 import checkedint.sticky : smartOp; // use IntFlagPolicy.sticky 2 3 assert(smartOp.mulPow2(-23, 5) == -736); 4 smartOp.mulPow2(10_000_000, 10); 5 assert(IntFlags.local.clear() == IntFlag.posOver); 6 7 assert(smartOp.mulPow2(65536, -8) == 256); 8 assert(smartOp.mulPow2(-100, -100) == 0);
Equivalent to left * pow(2, exp), but faster and works with a wider range of inputs. This is a safer alternative to left << exp that is still very fast.
Note that (conceptually) rounding occurs after the *, meaning that mulPow2(left, -exp) is equivalent to divPow2(left, exp).