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Haskell 夹中的计算步骤
原标题:Calculation steps in Haskell foldr
  • 时间:2010-08-20 08:52:53
  •  标签:
  • haskell
  • fold

是否有任何人知道有条不紊地利用职能的步骤?

GHCI Command Window:

foldr (x y -> 2*x + y) 4 [5,6,7] 

The result after evaluation:

40

Steps on this,

Prelude> foldr (x y -> 2*x + y) 4 [5,6,7] 
6 * 2 + (7 * 2 + 4)
12 + 18 = 30
5 * 2 + 30 = 40 v
最佳回答
问题回答

您实际上可以很容易地将其视而不见:

import Text.Printf

showOp f = f (printf "(%s op %s)") "0" ["1","2","3"]

之后

Main> showOp foldr
"(1 op (2 op (3 op 0)))"
Main> showOp foldl
"(((0 op 1) op 2) op 3)"
Main> showOp scanl
["0","(0 op 1)","((0 op 1) op 2)","(((0 op 1) op 2) op 3)"]




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