On Jul 9, 2011, at 8:38 PM, Clark Grubb wrote: > This seems to be a bug. Here is the > Racket behavior and Haskell behavior > for comparison. > > ====================== > > $ racket > Welcome to Racket v5.1.1. >> (foldl - 0 '(1 2 3))
John's stepper should show this: == (fold - (- 1 0) '(2 3)) == (fold - (- 2 1) '(3)) == (fold - (- 3 1) '()) == 2 Did you post the same message to the Haskell mailing group? Racket's fold heritage is significantly older than Haskell's. Is it perhaps possible that Haskell got it wrong? -- Matthias > 2 >> (foldr - 0 '(1 2 3)) > 2 > > ====================== > > $ ghci > GHCi, version 6.10.4: http://www.haskell.org/ghc/ :? for help > Prelude> foldl (-) 0 [1,2,3] > -6 > Prelude> foldr (-) 0 [1,2,3] > 2 > _________________________________________________ > For list-related administrative tasks: > http://lists.racket-lang.org/listinfo/dev _________________________________________________ For list-related administrative tasks: http://lists.racket-lang.org/listinfo/dev
