如何iterate修改功能以使结果为
f x, (f^2)x, (f^4)x, (f^8)x, ...
如果有人可以向我提供任何建议，我将非常高兴。
user721588
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2 回答
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Given, that f^x means f x-times applied to x I would say
iterate :: (a -> a) -> a -> [a]
iterate f x = f x : iterate (f . f) x
would suffice.
于 2012-06-11T18:25:36.673 回答
2
选择：
Prelude> map snd $ iterate (\(f,x) -> (f.f, f x)) ((+1),1)
[1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,...
于 2012-06-11T20:10:02.400 回答