Optimize pow when self is a power of two

if base == 2 ** k, then (2 ** k) ** n == 2 ** (k * n) == 1 << (k * n)

Author: Kmeakin

library/core/src/num/uint_macros.rs

@@ -3318,6 +3318,16 @@ macro_rules! uint_impl {
  let mut acc = 1;
 
  if intrinsics::is_val_statically_known(exp) {
+ // change of base:
+ // if base == 2 ** k, then
+ // (2 ** k) ** n
+ // == 2 ** (k * n)
+ // == 1 << (k * n)
+ if self.is_power_of_two() {
+ let k = self.ilog2();
+ return 1 << (k * exp)
+ }
+
  while exp > 1 {
  if (exp & 1) == 1 {
  acc = acc * base;