升压C ++ - 生成0和1之间的随机的实数 [英] Boost C++ - generating a random real number between 0 and 1

问题描述

我一直在阅读了Boost C ++文档，并试图找出如何产生0和1之间的随机实数，使用该库的 uniform_01 部分。有没有人有什么建议？

I've been reading the Boost C++ documentation and trying to figure out how to generate a random real number between 0 and 1, using the uniform_01 part of the library. Does anyone have any suggestions?

根据建议，我用这code，但它会产生相同的随机值每一次。

Based on suggestions, I'm using this code, but it generates the same random value every time.

double random01(mt19937 & generator)
{
    uniform_01<mt19937> dist(generator);
    return dist();
}

int main()
{
    mt19937 generator(time(0));
    for (int i = 0; i < 10; i++) {
        cout << random01(generator) << endl;
    }
    return 0;
}

很抱歉，如果我不清晰;这个问题不解决，我的答案或任何其他。它的每一次仍然产生相同的随机值。

Sorry if I wasn't clear; this question isn't solved, by my answer or any other. It's still generating the same random value each time.

推荐答案

有一些疑难杂症在这code：

There are a few 'gotcha's in this code:

1. 的 random01（）功能并不需要到发电机的引用，而是复制发电机。所以，如果你调用该函数使用相同的发电机多次 - 它会一遍又一遍产生相同的值


2. 的静态 uniform_01 在 random01（）只被初始化的第一次调用random01，所以如果是曾经与不同的生成调用，它不会使用它，而是使用第一发生器，它甚至可能已被破坏的参考，因为！

1. The random01() function does not take a reference to the generator, but rather a copy of the generator. So if you call the function multiple times with the same generator - it will produce the same value over and over again!
2. The static uniform_01 in random01() only gets initialized on the first call to random01, so if it is ever called with a different generator, it will not use it, but rather use a reference to the first generator, which may even have been destroyed, since!

正确的将类似以下内容（注意＆功放;在参数列表，用于传递按引用，而缺乏的静态：

Correct would be something like following (note the & in the arguments list, for pass-by-reference, and the lack of static:

double random01(mt19937 & generator)
{
    uniform_01<mt19937> dist(generator);
    return dist();
}

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