首页 \ 问答 \ 累加器如何在Haskell中工作?(How do accumulators work in Haskell?)

累加器如何在Haskell中工作?(How do accumulators work in Haskell?)

你好我试图通过创建以下方法来了解累加器如何工作:

它接收一个数字并创建一个元组:( (sum [0..nr], [0..nr]) 。 例如,给定4 ,它应该返回(10, [0,1,2,3,4])

这是我的代码:

accu :: Int -> (Int, [Int])
accu 0  = (0, [0])
accu nr = go (0, []) nr where   --I do not get this line, mainly the first argument
            go (s,ls) 0  = (s, 0:ls)
            go (s,ls) nr = go (s+nr, nr:ls) (nr-1)

我不明白累加器如何改变它的值,因为它不是一个函数,因此它不应该是不可变的? 在我的情况下,它不应该像一个常数吗?


Hello i am trying to understand how do accumulators work by creating the following method :

It receives a number and it creates a tuple of: (sum [0..nr], [0..nr]). For instance, given 4, it should return (10, [0,1,2,3,4]).

This is the code I have:

accu :: Int -> (Int, [Int])
accu 0  = (0, [0])
accu nr = go (0, []) nr where   --I do not get this line, mainly the first argument
            go (s,ls) 0  = (s, 0:ls)
            go (s,ls) nr = go (s+nr, nr:ls) (nr-1)

I do not understand how the accumulator changes its value, since it is not a function, and thus shouldn't it be immutable? Shouldn't it be like a constant in my case?


原文:https://stackoverflow.com/questions/49444631
更新时间:2019-12-13 07:41

最满意答案

go是一个使用where关键字在本地定义的函数,然后直接在accu函数中使用。 看看这里: http//learnyouahaskell.com/syntax-in-functions#where

这意味着累加器并没有真正改变,它只是每次递归调用go函数时的不同累加器。 最后一行将使用新的累加器值递归调用go函数。


go is a function that is locally defined using the where keyword and then used directly in the accu function. Have a look here: http://learnyouahaskell.com/syntax-in-functions#where

That means the accumulator does not really change, it is just a different accumulator in each recursive call to the go function. The last line will recursively call the go function with a new accumulator value.

2018-03-23

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