Hello Norman,
Friday, December 22, 2006, 8:23:57 AM, you wrote:
> compose :: (b -> c) -> (a -> b) -> (a -> c)
> compose f g = \x -> f (g x)
ghc 6.6 added 'inline' function, see user docs. although only SPJ knows
whether it can be used here:
compose f g = inline (\x -> f (g x))
--
Best regards
| My example is complicated, so let me present a simpler analogy.
| Suppose I defined
|
| compose :: (b -> c) -> (a -> b) -> (a -> c)
| compose f g = \x -> f (g x)
|
| I can easily persuade GHC to inline 'compose'.
| But when 'compose' is applied to known arguments, I wish
| f and g to be inlined i
I've just discovered the {-# INLINE #-} pragma, but it's not
doing as much for me as I had hoped.
My example is complicated, so let me present a simpler analogy.
Suppose I defined
compose :: (b -> c) -> (a -> b) -> (a -> c)
compose f g = \x -> f (g x)
I can easily persuade GHC to inline 'compose
