On Wed, Aug 18, 2010 at 4:38 PM, Lennart Augustsson < [email protected]> wrote:
> You don't know that f is strict in its first argument so you cannot > deduce that go is strict in z in the first case. > I'm not sure I understand. f :: Int -> Int -> Int -> Int f = \x y z -> x + y + z in this particular case so it's strict in all its parameters since + is strict (for Ints). f is inlined into go so it should be possible to deduce that go is indeed strict. As I mentioned to Ian you get the same result if you inline f manually. Cheers, Johan
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