使用uniform_int_distribution与模数运算有什么好处？

Question

根据以下结果,使用%操作在两个数字之间生成均匀的随机整数几乎比使用的快3倍std::uniform_int_distribution:是否有充分的理由使用std::uniform_int_distribution？

码:

#include <iostream>
#include <functional>
#include <vector>
#include <algorithm>
#include <random>

#include <cstdio>
#include <cstdlib>

using namespace std;

#define N 100000000

int main()
{

clock_t tic,toc;

for(int trials=0; trials<3; trials++)
{
    cout<<"trial: "<<trials<<endl;

    // uniform_int_distribution
    {
        int res = 0;
        mt19937 gen(1);
        uniform_int_distribution<int> dist(0,999);

        tic = clock();
        for(int i=0; i<N; i++)
        {
            int r = dist(gen);
            res += r;
            res %= 1000;
        }
        toc = clock();
        cout << "uniform_int_distribution: "<<(float)(toc-tic)/CLOCKS_PER_SEC << endl;
        cout<<res<<" "<<endl;

    }

    // simple modulus operation
    {
        int res = 0;
        mt19937 gen(1);

        tic = clock();
        for(int i=0; i<N; i++)
        {
            int r = gen()%1000;
            res += r;
            res %= 1000;
        }
        toc = clock();
        cout << "simple modulus operation: "<<(float)(toc-tic)/CLOCKS_PER_SEC << endl;
        cout<<res<<" "<<endl;

    }

    cout<<endl;
}

}

输出:

trial: 0
uniform_int_distribution: 2.90289
538 
simple modulus operation: 1.0232
575 

trial: 1
uniform_int_distribution: 2.86416
538 
simple modulus operation: 1.01866
575 

trial: 2
uniform_int_distribution: 2.94309
538 
simple modulus operation: 1.01809
575 

Answer 1

当您使用modulo()将范围映射到另一个区间时,您将获得统计偏差.%rand()

例如,假设rand()地图统一(没有偏见)[0, 32767],你想映射到[0,4]做rand() % 5.那么值0,1和2平均将在32768次中产生6554,但是值3和4仅产生6553次(因此3*6554 + 2*6553 = 32768).

偏差很小(0.01%),但取决于您的应用可能会致命.观看Stephan T. Lavavej的谈话" rand()被认为是有害的 "了解更多细节.