将函数映射到haskell中的列表/数组? [英] Mapping function to list/array in haskell?

问题描述

我正在尝试启动随机数组以开始游戏.为此，我创建了一个函数randomBoard，该函数返回'*'或''代表游戏板上的空间.

I am trying to initiate a random array to begin the game. To do that I create a function randomBoard which returns '*' or ' ' to represent a space on the game board.

我希望能够使用此功能创建游戏板阵列.我还没有成功实例化一个数组.我希望有一种方法可以声明一个大小为100的数组，并使用我的随机函数设置每个元素.

I want to be able to create the game board array with this function. I havent been able to succesfully instantiate an array yet. I am hoping there is a way i can declare an array of say size 100 and use my random function to set each element.

这显然行不通甚至无法编译.我肯定这有一些问题，因为我不确定如何在Haskell中甚至与IO一起工作并产生这种结果.任何指导都将不胜感激...

This obviously doesnt work or even compile. I am sure there are a couple things wrong with it, as I am not sure really how to even work with IO in haskell and produce this outcome. Any guidance is much appreciated...

推荐答案

这应该对您有用:

import Control.Monad
import System.Random
import Data.Array
import Data.List

randomBoard :: IO Char
randomBoard =
   do
   f1 <- randomIO :: IO Int
   if(f1 `mod` 2) == 0
     then return  '*'
     else return  ' '

boardArray :: IO (Array Int Char)
boardArray = listArray (0, 99) <$> replicateM 100 randomBoard

这就是我更改的内容:

1. 为清楚起见，我添加了类型签名randomBoard :: IO Char. (如果没有它，代码仍然可以工作，因为如果您不提供它，Haskell会正确推断出这种类型.)
2. 我将boardArray的类型更改为使用IO.任何使用IO的东西，无论多么间接，都必须存在于IO本身.
3. 我将listArray (0, 100)更改为listArray (0, 99)，因为前者实际上是101个元素.
4. map randomBoard $ 100 (0,100)根本不正确.要获取多个相同事物的列表，通常使用replicate，但是由于您在此关心的事物位于IO monad中，因此请改为使用replicateM. replicateM 100 randomBoard给出一个IO [Char]，其中包含100个'*'或' '随机元素.
5. 我添加了Control.Monad的导入，使用replicateM需要此导入.
6. 我在boardArray中使用<$>.由于要使用[Char]调用listArray并获得Array Int Char，但是replicateM 100 randomBoard是IO [Char]，因此不能直接应用该参数.使用<$>(也称为fmap)将其应用到IO的内部"，为您提供IO (Array Int Char).

1. I added the type signature randomBoard :: IO Char for clarity. (The code would still work without it, as Haskell correctly infers this type if you don't supply it.)
2. I changed the type of boardArray to use IO. Anything that uses IO, no matter how indirectly, need to be in IO itself.
3. I changed listArray (0, 100) to listArray (0, 99), as the former would actually be 101 elements.
4. map randomBoard $ 100 (0,100) isn't right at all. To get a list of several of the same thing, you'd usually use replicate, but since the thing you care about here is in the IO monad, you use replicateM instead. replicateM 100 randomBoard gives an IO [Char] with 100 random elements of either '*' or ' '.
5. I added an import of Control.Monad, which is needed to use replicateM.
6. I use <$> in boardArray. Since you want to call listArray with a [Char] and get an Array Int Char, but replicateM 100 randomBoard is an IO [Char], you can't just apply the argument directly. Using <$> (which is also called fmap) applies it "inside" of IO, giving you back an IO (Array Int Char).

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