On Nov 29, 2007 4:23 AM, PR Stanley
PRS: You would also get different results - e.g. let a = 3, b = 7, c = 2 therefore 20 = strict ( ( (a+(b*c)) ) therefore 17 = non-strict ( (a+(b*c)) )
or am I misunderstanding the concept?
Yes. If the strict program does not error, then the strict program and the lazy program will have the same results. Numerics are not the best way to illustrate the difference, because they are essentially strict in their semantics. How about a list function: head [] = error "empty list" head (x:xs) = x map f [] = [] map f (x:xs) = f x:map f xs head (map (+1) [1,2,3]) -- rewrite as... head (map (+1) (1:2:3:[])) Strictly would go like this: head (map (+1) (1:2:3:[])) -- evaluate map (+1) (1:2:3:[]) head ((1+1) : map (+1) (2:3:[])) -- evaluate 1+1 head (2 : map (+1) (2:3:[])) -- evaluate map (+1) (2:3:[]) head (2 : (2+1) : map (+1) (3:[])) -- evaluate 2+1 head (2 : 3 : map (+1) (3:[])) -- evaluate map (+1) (3:[]) head (2 : 3 : (3+1) : []) -- evaluate 3+1 head (2 : 3 : 4 : []) -- evaluate [] (nothing to do) head (2 : 3 : 4 : []) -- evaluate head 2 Lazily would go like this: head (map (+1) (1:2:3:[])) -- evaluate head -- try to match map (+1) (1:2:3:[]) -- against x:xs, need to evaluate map head ((1+1) : map (+1) (2:3:[])) -- evaluate head -- match (1+1):map (+1) (2:3:[]) against -- x:xs succeeds, with x = (1+1) (1+1) -- evaluate (1+1) 2 Here I'm describing lazy evaluation rather than non-strict semantics, but they're pretty closely related. Luke