[Haskell-cafe] foldl in terms of foldr
Eduard.Sergeev at gmail.com
Tue Jan 26 11:34:41 EST 2010
Eduard Sergeev wrote:
> 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).
Correction :) 4-arg and 3-arg respectively.
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