[英]How does Haskell evaluate this function which undoes list intercalation?
我试图了解Haskell如何评估sep [1, 2, 3, 4, 5]
以获得([1, 3], [2, 4, 5])
,其中:
sep [ ] = ([ ], [ ])
sep [x] = ([ ], [x])
sep (x1:x2:xs) = let (is, ps) = sep xs in (x1:is, x2:ps)
我开始是这样的:
sep [1, 2, 3, 4, 5] = let (is, ps) = sep [3, 4, 5] in (1:is, 2:ps)
但是之后?
终于我明白了。
1) sep [1, 2, 3, 4, 5] = let (is, ps) = sep [3, 4, 5] in (1:is, 2:ps)
2) sep [3, 4, 5] = let (is, ps) = sep [5] in (3:is, 4:ps)
3) sep [5] = ([], [5])
在2)中sep [3, 4, 5] = let (is, ps) = ([], [5]) in (3:is, 4:ps) = ([3], [4, 5])
在1)中sep [1, 2, 3, 4, 5] = let (is, ps) = ([3], [4, 5]) in (1:is, 2:ps) = ([1, 3], [2, 4, 5])
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