[Haskell-cafe] foldl in terms of foldr

Eduard Sergeev 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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