Re: [Haskell-cafe] Mystery operator?

Consider these two functions: z = ([], []) alt1 = foldr f z where f a (x,y) = (a:y, x) alt2 = foldr g z where g a ~(x,y) = (a:y, x) alt1 (1:2:3:undefined) = foldr f z (1:2:3:undefined) = f 1 (foldr f z (2:3:undefined)) -- Now f 1 needs to evaluate its second argument for the pattern match = f 1 (f 2 (foldr f z (3:undefined))) -- ditto for f 2 = f 1 (f 2 (f 3 (foldr f z undefined))) -- ditto for f 3 = f 1 (f 2 (f 3 undefined)) = undefined If you replace "undefined" with [], at this point we would start being able to return results. But notice how we had to evaluate the entire list before we can output any data at all. alt2 (1:2:3:undefined) = foldr g z (1:2:3:undefined) = g 1 (foldr g z (2:3:undefined)) -- g is a lazy pattern match for the 2nd argument = let (x,y) = foldr g z (2:3:undefined) in (1:y, x) Here we've already output a pair and the first element of the output, before we have even evaluated the tail of the list! Now if we continue demanding the output... = let (x,y) = g 2 (foldr g z (3:undefined)) in (1:y, x) = let (x1,y1) = foldr g z (3:undefined) (x,y) = (2:y1, x1) in (1:y,x) = let (x1,y1) = foldr g z (3:undefined) x = 2:y1 y = x1 in (1:y,x) = let (x1,y1) = foldr g z (3:undefined) in (1:x1,2:y1) = let (x1,y1) = g 3 (foldr g z undefined) in (1:x1, 2:y1) = let (x2,y2) = foldr g z undefined (x1,y1) = (3:y2, x2) in (1:x1, 2:y1) = let (x2,y2) = foldr g z undefined x1 = 3:y2 y1 = x2 in (1:x1, 2:y1) = let (x2,y2) = foldr g z undefined in (1:3:y2, 2:x2) = let (x2,y2) = undefined in (1:3:y2, 2:x2) = (1:3:undefined, 2:undefined) The lazy pattern match is allowing us to output as much of the result as possible; this implementation is "maximally lazy". As for guidelines: - Only use lazy pattern matching on single-constructor data types. Tuples are the prime candidate here. - Lazy pattern matches involve allocating additional thunks for the variables, so you are paying a cost. If you are immediately going to demand the contents of those variables (like in your 'add' example), there's no point. -- ryan On Mon, Nov 30, 2009 at 11:29 AM, David Menendez wrote:

On Mon, Nov 30, 2009 at 2:01 PM, michael rice wrote:

...

Hi all,

A lot of things posted here I wasn't aware of. My original example involved ~(x,y), so, returning to that context, how would these two simple cases vary:

add2 :: (Int,Int) -> Int add2 (x,y) = x+y

add2 :: (Int,Int) -> Int add2 ~(x,y) = x+y

I guess what I'm looking for is the concept that would dictate choosing one over the other.

These particular two functions are identical, because (+) is strict for Ints.

But consider these two functions:

swap ~(x,y) = (y,x) swap' (x,y) = (y,x)

swap is non-strict, and swap' is strict.

swap _|_ = (_|_, _|_) swap' _|_ = _|_

-- Dave Menendez http://www.eyrie.org/~zednenem/ _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe