Then if you turn :
fs True = True
fs x = True
to:
fs x = case x of
True -> True
x' -> True
Is it still strict, or does 'fs' wrap the case test and defer evaluation?
2011/4/1
Daniel Fischer wrote:
If you have a strict function, you may evaluate its argument eagerly without changing the result^1, while eager evaluation of a non-strict function's argument may produce _|_ where deferred evaluation wouldn't.
Sadly, that is quite untrue. Strictness is observable, already in Haskell98. That distressing result has nothing to do with imprecise exceptions, seq, non-termination, lack of resources, or the use of unsafe features. Plainly, just by changing the strictness of a function one may cause the program to print different results, such as "strict" or "non-strict" in the code below.
-- Haskell98! -- Strictness is observable -- With no seq, no unsafe operations
import Control.Exception
-- fs and fns are both essentially (const True) functions, -- but differ in strictness fs,fns :: Bool -> Bool
-- non-strict fns x = True
-- strict fs True = True fs x = True
handler :: SomeException -> IO () handler _ = print "strict"
test f = handle handler $ if f (error "Bang!") then print "non-strict" else return ()
main_s = test fs -- prints "strict"
main_ns = test fns -- prints "non-strict"
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