一.原理：
C語言中偽隨機數(shù)生成算法實際上是采用了"線性同余法"。具體的計算如下：?
seed = (seed * A + C ) % M
其中A,C,M都是常數(shù)（一般會取質(zhì)數(shù)）。當C=0時，叫做乘同余法。
假設(shè)我們定義隨機數(shù)函數(shù)：

void?rand(int?&seed)
{
????seed?=?(seed?*?A?+?C?)?%?M;
}

每次調(diào)用rand函數(shù)都會產(chǎn)生一個隨機值賦值給seed，可以看出實際上用rand函數(shù)生成的是一個遞推的序列，一切值都來源于最初的seed。所以當初始的seed取一樣的時候，得到的序列都相同。

我們稱seed為種子，一個偽隨機數(shù)常用的原則就是M盡可能的大。例如，對于32位的機器來說，選擇M=2^31-1=2147483647, A=7^5=16807時可以取得最佳效果。

二.代碼實現(xiàn)：
現(xiàn)在我們來看看levelDB里隨機數(shù)Random類是如何實現(xiàn)的：

在Random類中，A為16807，M為2147483647，C為0；

#ifndef?STORAGE_LEVELDB_UTIL_RANDOM_H_
#define?STORAGE_LEVELDB_UTIL_RANDOM_H_


#includenamespace?leveldb?{
????
????//?A?very?simple?random?number?generator.??Not?especially?good?at
????//?generating?truly?random?bits,?but?good?enough?for?our?needs?in?this
????//?package.
????
????class?Random
????{
????private:
????????uint32_t?seed_;
????public:
????????//?0x7fffffffu?==?2147483647L?==?2^31-1?==?01111111?11111111?11111111?11111111
????????//?表達式s?&?0x7fffffffu，確保結(jié)果值在[0,2147483647]范圍內(nèi)
????????explicit?Random(uint32_t?s)?:?seed_(s?&?0x7fffffffu)??
????????{
????????????//?Avoid?bad?seeds.
????????????//?seed_不能為零或M，否則所有的后續(xù)計算的值將為零或M
????????????if?(seed_?==?0?||?seed_?==?2147483647L)
????????????{
????????????????seed_?=?1;
????????????}
????????}
????????//?16807隨機數(shù)
????????uint32_t?Next()
????????{?
????????????//01111111?11111111?11111111?11111111
????????????static?const?uint32_t?M?=?2147483647L;???//?2^31-1
????????????//0100?0001?1010?0111
????????????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]
????????????//?這里將seed_*A設(shè)置為隨機數(shù)生成器product，注意到product是64位的。
????????????//?那么seed_=product?%?M就相當于得到大小在[1,M-1]之間的隨機數(shù)。
????????????uint64_t?product?=?seed_?*?A;
????????????//?Compute?(product?%?M)?using?the?fact?that?((x?<<?31)?%?M)?==?x.
????????????//?對于product?%?M，使用(product?>>?31)?+?(product?&?M)進行運算優(yōu)化，
????????????//?考慮到右移和與操作的代價遠小于取余操作。
????????????//?下面等式用到了((x?<<?31)?%?M)?==?x的技巧（等式的證明見第三節(jié)）
????????????//?product%M?==?static_cast((product?>>?31)?+?(product?&?M))的證明見第四節(jié)
????????????seed_?=?static_cast((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_;
????????}
????????//?Returns?a?uniformly?distributed?value?in?the?range?[0..n-1]
????????//?返回范圍[0..n-1]內(nèi)的均勻分布值。
????????//?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.
????????//?偏態(tài)：Uniform(max_log?+?1)取值范圍是[0,max_log]，1左移[0,max_log]得到
????????//?范圍是[1,2^max_log]，uniform（[1,2^max_log]）得到的范圍是[0,2^max_log-1]
????????uint32_t?Skewed(int?max_log)
????????{
????????????return?Uniform(1?<<?Uniform(max_log?+?1));
????????}
????};
????
}??//?namespace?leveldb

三.證明等式(x<<31)%M == x成立。其中M等于2^31-1
計算表達式左邊(x << 31) % M，由于x<<31等于x*2^31,
則(x << 31) % M=(x*2^31)%M=(x + x*(2^31-1))%M=(x + x*M)%M=x

四.證明等式(product%M) == (product>>31)+(product&M)，其中M等于2^31-1
因為product類型是uint64_t，可以將product從左到右分解成高33位和低31位，如下：
? ? ? ?高33位 ? ? ? ? ? ? ? ? ? ? ?低31
(product>>31)<<31+product&M
(product>>31)<