Haskell - 导致列表理解立即终止

Question

我想编写一个函数,有效地告诉我,当列表中的项目是在一个范围内的所有项目可分与否.问题是,我不知道在遇到false时立即停止列表理解.

allDivisByRng n start stop =
  [n `mod` x == 0 | x <- [start, start+1 .. stop]]

我已经尝试过,takeWhile但却无法做到我想做的事情.请指教.

allDivisByRng n start stop =
  takeWhile( == True )     [ n  `mod`  x == 0 | x <- [start, start+1.. stop] ]

当我跑步时,allDivisByRng 10 2 4我得到了一份真实的名单

其他一些样本运行

*Main> allDivisByRng 12 2 10
[True,True,True]
*Main> allDivisByRng 12 2 150
[True,True,True]
-- I want this to return false
*Main> allDivisByRng 2520 2 10
[True,True,True,True,True,True,True,True,True]
-- I want this to return True

我也试过寻找 False

allDivisByRng n start stop =
   takeWhile( == False )     [ n  `mod`  x == 0 | x <- [start, start+1.. stop] ]

*Main> allDivisByRng 10 2 3
[]
-- want false to be returned
*Main> allDivisByRng 10 2 10
[]
-- want false to be returned
*Main> allDivisByRng 2520 2 10
[]
-- want true to be returned
*Main> allDivisByRng 12 2 10
[]
-- want false to be returned 

Answer 1

功能all,

all :: Foldable t => (a -> Bool) -> t a -> Bool

允许您检查可折叠结构的所有元素是否满足特定条件.在你的情况下,

allDivisByRng n start stop = all ((== 0) . (mod n)) [start..stop]