[Haskell-cafe] foldl in terms of foldr

[Eduard Sergeev]

Xingzhi Pan wrote:
> 
> More over, does "foldr step f id xs z" equal to "foldr (step f) id xs z"??
> 

No, it does not. The former passes three-argument function 'step' to foldr
and the later passes two-argument function which is the result of the
partial application (step f).
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