在Haskell范围内生成无种子的随机整数 [英] Generate a random integer in a range in Haskell without a seed

问题描述

如何在不使用任何种子的情况下从范围(a，b)中在Haskell中生成随机数?

How can I generate a random number in Haskell from a range (a, b) without using any seed?

该函数应该返回一个Int而不是一个IO Int. 我有一个函数X，它带有Int和其他参数，并输出不是IO的东西.

The function should return an Int and not an IO Int. I have a function X that takes and Int and other arguments and outputs something which is not an IO.

如果这不可能，那么如何使用时间库生成种子并使用mkStdGen生成范围内的随机数?

If this is not possible, how can I generate a seed using the Time library and the generate a random number in the range with the mkStdGen ?

任何帮助将不胜感激.

推荐答案

一个函数不能在没有IO的情况下返回Int，除非它是一个纯函数，即给定相同的输入，您将始终获得相同的输出.这意味着，如果您想要一个不带IO的随机数，则将需要一个种子作为参数.

A function cannot return an Int without IO, unless it is a pure function, i.e. given the same input you will always get the same output. This means that if you want a random number without IO, you will need to take a seed as an argument.

• 如果选择获取种子，则种子的类型应为StdGen，并且可以使用randomR从中生成数字.使用newStdGen创建一个新种子(这必须在IO中完成).

• If you choose to take a seed, it should be of type StdGen, and you can use randomR to generate a number from it. Use newStdGen to create a new seed (this will have to be done in IO).

> import System.Random
> g <- newStdGen
> randomR (1, 10) g
(1,1012529354 2147442707)

randomR的结果是一个元组，其中第一个元素是随机值，第二个是用于生成更多值的新种子.

The result of randomR is a tuple where the first element is the random value, and the second is a new seed to use for generating more values.

否则，您可以使用randomRIO直接在IO monad中获取一个随机数，并为您提供所有StdGen内容:

Otherwise, you can use randomRIO to get a random number directly in the IO monad, with all the StdGen stuff taken care of for you:

> import System.Random
> randomRIO (1, 10)
6

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