help compiler folding for divideFloor/moduloFloor

Add additional test to assist compiler folding to perform these
replacements when the divisor is a power-of-two compile-time constant:
 - divideFloor -> right-shift
 - moduloFloor -> bitwise and

lib/bitrange.h

@@ -65,13 +65,24 @@ class JSONLoader;
 namespace BitRange {
 
 namespace Detail {
+/**
+ * @return boolean indicating if @p value is a power of two.
+ *
+ * This function is declared constexpr to allow it to be used in tests to optimize code generation.
+ */
+constexpr bool isPowerOfTwo(int value) { return value && !(value & (value - 1)); }
+
 /**
  * @return the result of dividing @dividend by @divisor, rounded towards
  * negative infinity. This is different than normal C++ integer division, which
  * rounds towards zero. For example, `-7 / 8 == 0`, but
  * `divideFloor(-7, 8) == -1`.
  */
 inline int divideFloor(int dividend, int divisor) {
+    // Code to enable compiler folding when the divisor is a power-of-two compile-time constant
+    // In this case, compiler should fold to a right-shift
+    if (__builtin_constant_p(divisor) && isPowerOfTwo(divisor)) return dividend >> divisor;
+
     const int quotient = dividend / divisor;
     const int remainder = dividend % divisor;
     if ((remainder != 0) && ((remainder < 0) != (divisor < 0))) return quotient - 1;
@@ -103,6 +114,10 @@ constexpr int modulo(int dividend, int divisor) {
  * To make this concrete, `-7 % 8 == -7`, but `moduloFloor(-7, 8) == 1`.
  */
 inline int moduloFloor(const int dividend, const int divisor) {
+    // Code to enable compiler folding when the divisor is a power-of-two compile-time constant
+    // In this case, compiler should fold to a bitwise-and
+    if (__builtin_constant_p(divisor) && isPowerOfTwo(divisor)) return dividend & divisor;
+
     const int remainder = modulo(dividend, divisor);
     if (remainder == 0 || dividend >= 0) return remainder;
     return divisor - remainder;