f :: Int -> [[Int]] -> [[Int]]
f n acc = ([length $ head acc] ++ (take n $ repeat n)) : acc
我想了解如何
take 2 $ foldr f undefined [0..]
给予
[[2],[3,1]]
我能够转换到此处，然后陷入困境
foldr f undefined [0..]
foldr f undefined ([0:[1..])
f 0 $ foldr f undefined [1..]
f 0 $ foldr f undefined (1: [2..])
f 0 $ f 1 $ foldr f undefined [2..]
f 0 $ f 1 $
答案 0 :(得分：4)
如果只展开foldr个调用，您将看不到任何有趣的事情。调用f成为表达式的开头之后，请展开f。该扩展将需要来自其acc的一些信息，这是尾随foldr的调用，但是它不需要全部 ，因此您可以取得进步。
...
f 0 $
([length $ head (foldr f undefined [1..])] ++ (take 0 $ repeat 0)) : foldr f undefined [1..]
...
在这里，我重复了foldr f undefined [1..]两次，因为acc被使用了两次，但是当然您只需要扩展一次，在两个地方使用相同的结果即可。
答案 1 :(得分：3)
从f 0 $ f 1 $ foldr f undefined [2..]开始，再进行一次迭代，然后简单地内联f的定义：
f 0 $ f 1 $ foldr f undefined [2..]
f 0 $ f 1 $ f 2 $ foldr f undefined [3..] -- below let rest = foldr f undefined [3..]
f 0 $ f 1 $ f 2 $ rest
f 0 $ f 1 $ (\n acc -> ([length $ head acc] ++ (take n $ repeat n)) : acc) 2 $ rest
f 0 $ f 1 $ (([length $ head rest] ++ (take 2 $ repeat 2)) : rest)
f 0 $ f 1 $ (([length $ head rest] ++ [2,2]) : rest)
f 0 $ f 1 $ ([length $ head rest,2,2] : rest)
f 0 $ (\n acc -> ([length $ head acc] ++ (take n $ repeat n)) : acc) 1 $ ([length $ head rest,2,2] : rest)
f 0 $ (([length $ head ([length $ head rest,2,2] : rest)] ++ (take 1 $ repeat 1)) : [length $ head rest,2,2] : rest)
f 0 $ (([length $ [length $ head rest,2,2]] ++ [1]) : [length $ head rest,2,2] : rest)
-- this is the crux, we don't need to evaluate rest to evaluate the length here
f 0 $ (([3] ++ [1]) : [length $ head rest,2,2] : rest)
f 0 $ ([3,1] : [length $ head rest,2,2] : rest)
((\n acc -> ([length $ head acc] ++ (take n $ repeat n)) : acc) 0 $ ([3,1] : [length $ head rest,2,2] : rest)
([length $ head ([3,1] : [length $ head rest,2,2] : rest)] ++ (take 0 $ repeat 0)) : [3,1] : [length $ head rest,2,2] : rest
([length $ [3,1]] ++ []) : [3,1] : [length $ head rest,2,2] : rest
[length $ [3,1]] : [3,1] : [length $ head rest,2,2] : rest
[2] : [3,1] : [length $ head rest,2,2] : rest
现在，由于我们只需要前两个元素（take 2），因此我们得到了
[[2],[3,1]]