tee*_*jay 6 c++ random prng mersenne-twister c++11
问:如何从a-priory未知范围生成(多个)均匀分布的整数?在性能方面(生成的数字的数百万),首选的方式是什么?
上下文:在我的应用程序中,我必须在许多地方生成许多伪随机数.我使用单例模式生成器来保持应用程序运行的可重复性.在我的情况下,分布总是一致的,但问题是在C++ 11样式中预先制作分发对象有太多可能的范围.
我尝试过:有两个明显的解决方案,第一个是一次性分配对象,第二个是使用modulo将随机数从最宽的范围转换为所需的范围.但不知何故,我怀疑这些是最好的:)
#include <random>
#include <iostream>
#include "limits.h"
using namespace std;
mt19937 mt;
uniform_int_distribution<int> * fixedDist;
uniform_int_distribution<int> * variableDist;
// this version creates and delete dist after just one use
int getIntFromRange1(int from, int to){
variableDist = new uniform_int_distribution<int>(from,to);
int num = (*variableDist)(mt);
delete variableDist;
return num;
}
// this version contains modulo
int getIntFromRange2(int from, int to){
int num = (*fixedDist)(mt);
int diff = to - from;
num = num % diff;
return num + from;
}
int main(){
mt.seed(123456);
fixedDist= new uniform_int_distribution<int>(0,INT_MAX)
int a = getIntFromRange1(1,10); // 1 and 10 are just for illustration
int b = getIntFromRange2(1,10); // can change freely
cout << "a: " << a << endl; // a: 6
cout << "b: " << b << endl; // b: 9
getchar();
}
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Jar*_*d42 10
我会做
int getIntFromRange1(int from, int to){
std::uniform_int_distribution<int> dist(from, to);
return dist(mt);
}
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