Re: [Haskell-cafe] length defined with foldr

On Wed, 19 Sep 2007, Stefan O'Rear wrote:

On Thu, Sep 20, 2007 at 04:17:54AM +0100, PR Stanley wrote:

...

Hi length = foldr (. n . 1 + n) 0 Any idea how to define length with foldr. The above definition doesn't make much sense. Many thanks, Paul

or, in ASCII concrete syntax

length = foldr (\_ n -> 1 + n) 0

or length = foldr (const succ) 0 :-) (But why foldr and not foldl' ? Working with lazy Peano numbers?)