【LevelDB源码阅读】Random

是什么

用于产生随机数。
C语言中伪随机数生成算法实际上是采用了“线性同余法”，具体计算如下：

seed = (seed * A + C ) % M

其中A,C,M都是常数（一般取质数），当C=0时，叫作乘同余法。

为什么要用

为什么不用系统随机数？

学到什么

• 可以将长的二进制整数分解为多个段来解决问题
• 利用位运算优化算术运算

源码分析

构造函数
因为后面随机数生成采用seed_ = (seed_ * A) % M，如果seed_为0或M(2^31-1)会导致所有seed_都是0。

// 0x7fffffffu == 2147483647L == 2^31-1 == 01111111 11111111 11111111 11111111
  // 表达式s & 0x7fffffffu，确保结果值在[0,2147483647]范围内
  explicit Random(uint32_t s) : seed_(s & 0x7fffffffu) {
    // Avoid bad seeds.
    if (seed_ == 0 || seed_ == 2147483647L) {
      seed_ = 1;
    }
  }

Next方法
在计算produce%M时使用static_cast<uint32_t>((product >> 31) + (product & M))来进行优化，因为左移运算和与操作快于模操作。下面证明

produce%M == static_cast<uint32_t>((product >> 31) + (product & M))

product类型是uint64_t，可以将product二进制从左到右分为高33位和低31位，假设高33位为H，低31位为L，则：
product = H << 31 + L
左边 = produce % M = (H << 31 + L) % M = (H * 2^31 + L) % M=(H * M + H + L) % M = H + L
右边 = (product >> 31) + (product & M) = (H * 2^31 + L) >> 31 + L = (H * 2^31 + L) / 2^31 + L = H + L
因为低31位可能等于M，则左边 = H，右边 = H + L，所以需要判断当右边大于M时，减去M。

uint32_t Next() {
    static const uint32_t M = 2147483647L;  // 2^31-1
    static const uint64_t A = 16807;        // bits 14, 8, 7, 5, 2, 1, 0
    // We are computing
    //       seed_ = (seed_ * A) % M,    where M = 2^31-1
    //
    // seed_ must not be zero or M, or else all subsequent computed values
    // will be zero or M respectively.  For all other values, seed_ will end
    // up cycling through every number in [1,M-1]
    uint64_t product = seed_ * A;

    // Compute (product % M) using the fact that ((x << 31) % M) == x.
    seed_ = static_cast<uint32_t>((product >> 31) + (product & M));
    // The first reduction may overflow by 1 bit, so we may need to
    // repeat.  mod == M is not possible; using > allows the faster
    // sign-bit-based test.
    if (seed_ > M) {
      seed_ -= M;
    }
    return seed_;
  }

其它接口

// Returns a uniformly distributed value in the range [0..n-1]
  // REQUIRES: n > 0
  uint32_t Uniform(int n) { return Next() % n; }

  // Randomly returns true ~"1/n" of the time, and false otherwise.
  // REQUIRES: n > 0
  bool OneIn(int n) { return (Next() % n) == 0; }

  // Skewed: pick "base" uniformly from range [0,max_log] and then
  // return "base" random bits.  The effect is to pick a number in the
  // range [0,2^max_log-1] with exponential bias towards smaller numbers.
  uint32_t Skewed(int max_log) { return Uniform(1 << Uniform(max_log + 1)); }