leveldb在random.h中实现一个非常简单的随机数生成器，说成简陋也可以。
源码分析
构造函数
explicit Random(uint32_t s) : seed_(s & 0x7fffffffu) {
// Avoid bad seeds.
if (seed_ == 0 || seed_ == 2147483647L) {
seed_ = 1;
}
}
构造函数用来初始化随机种子数，注意到作者认为当seed_为0或者 M(2^31-1)时是一种不好的种子数。这是因为在后面的随机数生成方法中采用的是seed_ = (seed_ * A) % M,如果seed_为0或者M,那么以后产生的所有seed_都是0了(具体表现在Next()函数中)。
Next()
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_;
}
在Next()函数中，开始我没看明白的一点是seed_ 如何由product得来的，其实注意到product是一个uint64_t,而seed_和M是uint32_t。如同注释中提示到的Compute (product % M) using the fact that ((x << 31) % M) == x.,将(product % M)分成高32位的结果和低32位的结果，然后相加即可。(不知道为何能分开求的可以去面壁了)
Uniform() && OneIn()
// 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; }
非常简单的两个函数，注释也写的很明白，要是不明白。那也没办法了
Skewed()
// 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));
}
这个函数随机返回[0, 2^(max_log-1)]区间中的一个2的指数值。
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