C ++ 11:如何使用< random>设置种子 [英] C++11: How to set seed using <random>

问题描述

我正在练习C ++ 11的新随机库.我编写了以下最小程序:

I am exercising the random library, new to C++11. I wrote the following minimal program:

#include <iostream>
#include <random>
using namespace std;
int main() {
    default_random_engine eng;
    uniform_real_distribution<double> urd(0, 1);
    cout << "Uniform [0, 1): " << urd(eng);
}

当我重复运行此命令时，每次都会给出相同的输出:

When I run this repeatedly it gives the same output each time:

>a
Uniform [0, 1): 0.131538
>a
Uniform [0, 1): 0.131538
>a
Uniform [0, 1): 0.131538

我希望让程序在每次调用时都设置不同的种子，以便每次生成不同的随机数.我知道random提供了一种称为seed_seq的工具，但我发现它的解释(在cplusplus.com上)完全晦涩:

I would like to have the program set the seed differently each time it is called, so that a different random number is generated each time. I am aware that random provides a facility called seed_seq, but I find the explanation of it (at cplusplus.com) totally obscure:

http://www.cplusplus.com/reference/random/seed_seq/

对于在每次调用程序时如何生成新种子的建议，我表示赞赏:越简单越好.

I'd appreciate advice on how to have a program generate a new seed each time it is called: The simpler the better.

我的平台:

• Windows 7: TDM-GCC编译器

推荐答案

具有seed_seq的目的是增加所生成序列的熵.如果您的系统上具有random_device，则可以使用该随机设备中的多个数字进行初始化.在具有伪随机数生成器的系统上，我认为随机性(即生成的序列熵)不会增加.

The point of having a seed_seq is to increase the entropy of the generated sequence. If you have a random_device on your system, initializing with multiple numbers from that random device may arguably do that. On a system that has a pseudo-random number generator I don't think there is an increase in randomness, i.e. generated sequence entropy.

以您的方法为基础:

如果您的系统确实提供了随机设备，则可以这样使用它:

If your system does provide a random device then you can use it like this:

  std::random_device r;
  // std::seed_seq ssq{r()};
  // and then passing it to the engine does the same
  default_random_engine eng{r()};
  uniform_real_distribution<double> urd(0, 1);
  cout << "Uniform [0, 1): " << urd(eng);

如果您的系统没有随机设备，则可以使用time(0)作为random_engine的种子

If your system does not have a random device then you can use time(0) as a seed to the random_engine

  default_random_engine eng{static_cast<long unsigned int>(time(0))};
  uniform_real_distribution<double> urd(0, 1);
  cout << "Uniform [0, 1): " << urd(eng);

如果您有多个随机来源，则可以实际执行此操作(例如2个)

If you have multiple sources of randomness you can actually do this (e.g. 2)

std::seed_seq seed{ r1(), r2() };
  default_random_engine eng{seed};
  uniform_real_distribution<double> urd(0, 1);
  cout << "Uniform [0, 1): " << urd(eng);

其中r1，r2是不同的随机设备，例如热噪声或量子源.

where r1 , r2 are different random devices , e.g. a thermal noise or quantum source .

您当然可以混合搭配

std::seed_seq seed{ r1(), static_cast<long unsigned int>(time(0)) };
  default_random_engine eng{seed};
  uniform_real_distribution<double> urd(0, 1);
  cout << "Uniform [0, 1): " << urd(eng);

最后，我想使用一个衬里进行初始化:

Finally, I like to initialize with an one liner:

  auto rand = std::bind(std::uniform_real_distribution<double>{0,1},
              std::default_random_engine{std::random_device()()});
  std::cout << "Uniform [0,1): " << rand();

---

如果您担心time(0)具有第二精度，则可以通过使用high_resolution_clock来解决此问题，方法是请求

---

If you worry about the time(0) having second precision you can overcome this by playing with the high_resolution_clock either by requesting the time since epoch as designated firstly by bames23 below:

static_cast<long unsigned int>(std::chrono::high_resolution_clock::now().time_since_epoch().count()) 

或者只是玩CPU随机性

or maybe just play with CPU randomness

long unsigned int getseed(int const K)
{

    typedef std::chrono::high_resolution_clock hiclock;

    auto gett= [](std::chrono::time_point<hiclock> t0)
    {
        auto tn = hiclock::now();
        return static_cast<long unsigned int>(std::chrono::duration_cast<std::chrono::microseconds>(tn-t0).count());
    };

    long unsigned int diffs[10];
    diffs[0] = gett(hiclock::now());
    for(int i=1; i!=10; i++)
    {
        auto last = hiclock::now();
        for(int k=K; k!=0; k--)
        {
            diffs[i]= gett(last);
        }
    }

    return *std::max_element(&diffs[1],&diffs[9]);
}

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