/// \brief Add two unsigned integers, X and Y, of type T.
/// Clamp the result to the maximum representable value of T on overflow.
/// \brief Add two unsigned integers, X and Y, of type T.
/// Clamp the result to the maximum representable value of T on overflow.
-SaturatingAdd(T X, T Y) {
+SaturatingAdd(T X, T Y, bool *ResultOverflowed = nullptr) {
+ bool Dummy;
+ bool &Overflowed = ResultOverflowed ? *ResultOverflowed : Dummy;
/// \brief Multiply two unsigned integers, X and Y, of type T.
/// Clamp the result to the maximum representable value of T on overflow.
/// \brief Multiply two unsigned integers, X and Y, of type T.
/// Clamp the result to the maximum representable value of T on overflow.
-SaturatingMultiply(T X, T Y) {
+SaturatingMultiply(T X, T Y, bool *ResultOverflowed = nullptr) {
+ bool Dummy;
+ bool &Overflowed = ResultOverflowed ? *ResultOverflowed : Dummy;
+
// Hacker's Delight, p. 30 has a different algorithm, but we don't use that
// because it fails for uint16_t (where multiplication can have undefined
// behavior due to promotion to int), and requires a division in addition
// to the multiplication.
// Hacker's Delight, p. 30 has a different algorithm, but we don't use that
// because it fails for uint16_t (where multiplication can have undefined
// behavior due to promotion to int), and requires a division in addition
// to the multiplication.
// Log2(Z) would be either Log2Z or Log2Z + 1.
// Special case: if X or Y is 0, Log2_64 gives -1, and Log2Z
// will necessarily be less than Log2Max as desired.
int Log2Z = Log2_64(X) + Log2_64(Y);
const T Max = std::numeric_limits<T>::max();
int Log2Max = Log2_64(Max);
// Log2(Z) would be either Log2Z or Log2Z + 1.
// Special case: if X or Y is 0, Log2_64 gives -1, and Log2Z
// will necessarily be less than Log2Max as desired.
int Log2Z = Log2_64(X) + Log2_64(Y);
const T Max = std::numeric_limits<T>::max();
int Log2Max = Log2_64(Max);
// We're going to use the top bit, and maybe overflow one
// bit past it. Multiply all but the bottom bit then add
// that on at the end.
T Z = (X >> 1) * Y;
// We're going to use the top bit, and maybe overflow one
// bit past it. Multiply all but the bottom bit then add
// that on at the end.
T Z = (X >> 1) * Y;