std :: uniform_real_distribution包含范围 [英] std::uniform_real_distribution inclusive range

问题描述

C ++ 11 std :: uniform_real_distribution（-1，1）提供范围[-1,1）中的数字。

C++11 std::uniform_real_distribution( -1, 1 ) gives numbers in the range [-1,1).

真实分布范围[-1,1]？

How would you get a uniform real distribution in the range [-1,1]?

实际上它可能没有关系，但在逻辑上，我试图选择一个包含范围内的值。 / p>

Practically it probably doesn't matter but logically I'm trying to select a value in the inclusive range.

推荐答案

如果你从整数开始，这更容易想。如果你传递[-1，1），你会得到 -1,0 。由于您想要包含 1 ，您将传递[-1，（1 + 1））或[-1，2]。现在你得到 -1,0,1 。

This is easier to think about if you start by looking at integers. If you pass [-1, 1) you would expect to get -1, 0. Since you want to include 1, you would pass [-1, (1+1)), or [-1, 2). Now you get -1, 0, 1.

你想做同样的事情，

借用此答案：

#include <cfloat> // DBL_MAX
#include <cmath> // std::nextafter
#include <random>
#include <iostream>

int main()
{
  const double start = -1.0;
  const double stop = 1.0;

  std::random_device rd;
  std::mt19937 gen(rd());

  // Note: uniform_real_distribution does [start, stop),
  //   but we want to do [start, stop].
  //   Pass the next largest value instead.
  std::uniform_real_distribution<> dis(start, std::nextafter(stop, DBL_MAX));

  for (auto i = 0; i < 100; ++i)
  {
    std::cout << dis(gen) << "\n";
  }
  std::cout << std::endl;
}

（请参阅代码运行这里）

也就是说，找到下一个最大的double值，然后将其作为结束值。

That is, find the next largest double value after the one you want, and pass that as the end value instead.

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