使用C ++ 11随机库生成随机双精度数 [英] Generate random double numbers using C++11 random library

问题描述

正如标题所示，我试图找出一种使用新的c ++ 11 < random> 库生成随机数的方法。我已经尝试了这个代码：

As the title suggests, I am trying to figure out a way of generating random numbers using the new c++11 <random> library. I have tried it with this code:

std::default_random_engine generator;
std::uniform_real_distribution<double> uniform_distance(1, 10.001);

我的代码的问题是每次我编译和运行它，它总是生成相同的数字。所以我的问题是随机库中的其他函数可以实现这一点，而是真正随机的？

The problem with the code I have is that every time I compile and run it, it always generates the same numbers. So my question is what other functions in the random library can accomplish this while being truly random?

对于我的特定用例，我试图得到一个范围 [1，10]

For my particular use case, I was trying to get a range within [1, 10]

推荐答案

来自Microsoft的Stephan T. Lavavej在Going Native上谈了如何使用新的C ++ 11随机函数，为什么不使用 rand（）。在其中，他包括一个幻灯片，基本上解决了你的问题。我已从下面的幻灯片复制了代码。

Stephan T. Lavavej (stl) from Microsoft did a talk at Going Native about how to use the new C++11 random functions and why not to use rand(). In it, he included a slide that basically solves your question. I've copied the code from that slide below.

您可以在这里看到他的全部讨论：http://channel9.msdn.com/Events/GoingNative/2013/rand-Considered-有害

You can see his full talk here: http://channel9.msdn.com/Events/GoingNative/2013/rand-Considered-Harmful

#include <random>
#include <iostream>

int main() {
    std::random_device rd;
    std::mt19937 mt(rd());
    std::uniform_real_distribution<double> dist(1.0, 10.0);

    for (int i=0; i<16; ++i)
        std::cout << dist(mt) << "\n";
}

我们使用 random_device 一次到种子为 mt 的随机数生成器。 random_device（）慢于 mt19937 ，但不需要进行播种，因为它从您的操作系统（它将从各种位置来源，例如 RdRand ）。

We use random_device once to seed the random number generator named mt. random_device() is slower than mt19937, but it does not need to be seeded because it requests random data from your operating system (which will source from various locations, like RdRand for example).

查看此问题/答案，看起来 uniform_real_distribution 返回 [a，b）范围内的数字， code> [a，b] 。为此，我们的 uniform_real_distibution 实际上应该是：

Looking at this question / answer, it appears that uniform_real_distribution returns a number in the range [a, b), where you want [a, b]. To do that, our uniform_real_distibution should actually look like:

std::uniform_real_distribution<double> dist(1, std::nextafter(10, DBL_MAX));

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