在 Haskell 中循环时 [英] While in a loop in Haskell

问题描述

如何在 Haskell 中编写以下伪代码?

How to code the following pseudo-code in Haskell?

x=0
for (i from 0 to 100):
    j=0
    while ( f(i,j) >0 ):
      x+= f(i,j)
      j+=1

(f 一些不重要的函数.)

(f some unimportant function.)

我想出了这样的事情:

a= [x| i<-[0..100], let s = takeWhile (k-> (f i k > 0)) [0..],
        j<- s, let x = f i j ]

然后 Sum a 完成工作，但我需要计算 f i j 两次，这有点多余.

Then Sum a does the work, but I need to compute f i j two times which is a little bit redundant.

这是否可以通过 f 只计算一次或一些运行速度更快的更好的代码来完成?

Can this be done with f computed only once or some better codes that run faster?

推荐答案

这里有一种方法，每对只计算一次 f:

Here's a way that only computes f once for each pair:

inner i = sum $ takeWhile (> 0) $ map (f i) [0..]
x= sum $ map inner [0..100]

我不喜欢列表推导式，尤其是对于更复杂的表达式，所以我发现您的解决方案难以阅读.主要区别在于，不是存储 j 的列表，使得 f i j >0，我存储了实际的函数值.由于懒惰，这不再起作用.

I dislike list comprehensions, especially for more complex expressions, so I found your solution difficult to read. The primary difference is that instead of storing a list of js such that f i j > 0, I stored the actual function value. This does no more work due to laziness.

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