两个压缩有符号整数的乘法

Question

Stockfish 国际象棋引擎需要存储残局得分和中局得分，以便进行评估。

它将它们打包成一个int，而不是单独存储它们。中局得分存储在低 16 位中。残局得分存储在较高的 16 位中，如果中局得分为正则按原样存储，如果为负则为负一。

这样做的好处是可以对两个数字并行进行运算（加法、减法、求反法和乘法）。

Here is the code:

/// Score enum stores a middlegame and an endgame value in a single integer (enum).
/// The least significant 16 bits are used to store the middlegame value and the
/// upper 16 bits are used to store the endgame value. We have to take care to
/// avoid left-shifting a signed int to avoid undefined behavior.
enum Score : int { SCORE_ZERO };

constexpr Score make_score(int mg, int eg) {
  return Score((int)((unsigned int)eg << 16) + mg);
}

/// Extracting the signed lower and upper 16 bits is not so trivial because
/// according to the standard a simple cast to short is implementation defined
/// and so is a right shift of a signed integer.
inline Value eg_value(Score s) {
  union { uint16_t u; int16_t s; } eg = { uint16_t(unsigned(s + 0x8000) >> 16) };
  return Value(eg.s);
}

inline Value mg_value(Score s) {
  union { uint16_t u; int16_t s; } mg = { uint16_t(unsigned(s)) };
  return Value(mg.s);
}

#define ENABLE_BASE_OPERATORS_ON(T)                                \
constexpr T operator+(T d1, int d2) { return T(int(d1) + d2); }    \
constexpr T operator-(T d1, int d2) { return T(int(d1) - d2); }    \
constexpr T operator-(T d) { return T(-int(d)); }                  \
inline T& operator+=(T& d1, int d2) { return d1 = d1 + d2; }       \
inline T& operator-=(T& d1, int d2) { return d1 = d1 - d2; }

ENABLE_BASE_OPERATORS_ON(Score)

/// Only declared but not defined. We don't want to multiply two scores due to
/// a very high risk of overflow. So user should explicitly convert to integer.
Score operator*(Score, Score) = delete;

/// Division of a Score must be handled separately for each term
inline Score operator/(Score s, int i) {
  return make_score(mg_value(s) / i, eg_value(s) / i);
}

/// Multiplication of a Score by an integer. We check for overflow in debug mode.
inline Score operator*(Score s, int i) {

  Score result = Score(int(s) * i);

  assert(eg_value(result) == (i * eg_value(s)));
  assert(mg_value(result) == (i * mg_value(s)));
  assert((i == 0) || (result / i) == s);

  return result;
}

我了解加法、减法和求反法是如何工作的，但我难以理解的是乘法。将整数相乘如何正确地将残局和中局分数相乘？

Answer 1

将两个压缩整数相乘是行不通的。但是只要你有足够的填充位，乘以一个常数就可以了。

想象一下以十为基数，它变得更容易：

23和31包含一千个因素：230031

让我们乘以 3，我们将得到 69 和 93

还有230031 * 3是690093。因此，只要最低位的数字不会溢出最高位的数字，您就可以同时进行这两个乘法运算。

如果您以十六进制查看您的数字，同样的魔法也会发生。但是，对我来说，十进制更容易理解。