It's not too difficult to do: think 'fold map' and put it in the form that foldr needs.

cheers

-----Original Message-----
From: James Ealing [mailto:jamesealing2000@yahoo..co.uk]
Sent: Friday, 02 January, 2004 1:54 PM
To: haskell-cafe@haskell.org
Subject: foldr

Hi all,

If I have a function:

it f [a0, a1, a2, ...] = [a0, f a1, f (f a2), ....]

Is there any way of expressing

it f

as an instance of foldr?

Many thanks,

Jim

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