[Haskell-cafe] and from standard Prelude

[Oleg Lobachev]

Hello all,

the and function, and :: [Bool] -> Bool is defined in two different ways in the latest Prelude.

I would expect it to be

> and = foldr (&&) True

However, there is a further recursive definition, and it is the one used! See http://haskell.org/ghc/docs/6.12.2/html/libraries/base-4.2.0.1/src/GHC-List.html#and

> #ifdef USE_REPORT_PRELUDE
> and                     =  foldr (&&) True
> or                      =  foldr (||) False
> #else
> and []          =  True
> and (x:xs)      =  x && and xs
> or []           =  False
> or (x:xs)       =  x || or xs
> 
> {-# RULES
> 
> "and/build"     forall (g::forall b.(Bool->b->b)->b->b) . 
> 
>                 and (build g) = g (&&) True
>
> "or/build"      forall (g::forall b.(Bool->b->b)->b->b) . 
> 
>                or (build g) = g (||) False
> 
> #-}
> #endif

Is there any reason for that besides some clever optimizations? I am particularly interested in the behavior of and for infinite lists and these two versions are equivalent in this aspect.

Greetings,
Oleg