生成随机整数时无法包含上限

问题描述

我正在编写一个方法，它将生成一个 1 到 6 之间的无符号整数(包括边界).我目前的方法如下.

I am writing a method that will generate an unsigned int between 1 and 6 (boundaries included). The current method I have is below.

        private static Random random = new Random();
        ...
        private static uint GetRandomChannel()
        {
            return Convert.ToUInt32(random.Next(1, 6));
        }

我已经运行这个方法一千次了，我得到了数字 1 到 5，但从来没有得到过 6.为什么会发生这种情况，我该如何解决?

I've run this method a thousand times and I get numbers 1 through 5 but never get 6. Why is this happening and how can I fix it?

推荐答案

random.Next() 是一个独占上限.

参数

minValue:返回的随机数的包含下限.

minValue: The inclusive lower bound of the random number returned.

maxValue: 返回的随机数的唯一上限.maxValue 必须大于或等于 minValue.

maxValue: The exclusive upper bound of the random number returned. maxValue must be greater than or equal to minValue.

返回值

大于或等于 minValue 且小于 maxValue 的 32 位有符号整数；即返回值的范围包括minValue，但不包括maxValue.如果 minValue 等于 maxValue，则返回 minValue.

A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.

这意味着 random.Next(1, 6) 只会返回 n 范围内的值 1 .

This means that random.Next(1, 6) will only return values n in the range 1 <= n < 6.

因此对于您的模具滚动模拟，您需要使用

So for your die rolling simulation you will need to use

random.Next(1, 7)

注意:这个 API 的设计很奇怪.它对 minValue == maxValue 有特殊情况处理，这似乎不必要地使 API 复杂化.如果我设计了这个 API，我会把两个参数都设为包含限制.这将产生令人愉悦的对称性，并允许随机数覆盖 int 的全部范围.

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Note: The design of this API is odd. It has special case handling for minValue == maxValue which seems to needlessly complicate the API. If I had designed this API I would have made both parameters be inclusive limits. This would have resulted in a pleasing symmetry and would have allowed random numbers that cover the full range of int.

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