On Sun, Jun 28, 2009 at 10:05 PM, Tony Morris
Is there a canonical function for traversing the spine of a list?
I could use e.g. (seq . length) but this feels dirty, so I have foldl' (const . const $ ()) () which still doesn't feel right. What's the typical means of doing this?
You can also use the combinators in Control.Parallel.Strategies. 'seqList r0' will force the spine, but you also can swap in different strategies to force evaluation on the elements as well. Prelude Control.Parallel.Strategies> let b = [ [1..n] | n <- [1..5] ] Prelude Control.Parallel.Strategies> :print b b = (_t1::[[Integer]]) Prelude Control.Parallel.Strategies> seqList r0 b () Prelude Control.Parallel.Strategies> :print b b = [(_t2::[Integer]),(_t3::[Integer]),(_t4::[Integer]), (_t5::[Integer]),(_t6::[Integer])] Prelude Control.Parallel.Strategies> seqList rwhnf b () Prelude Control.Parallel.Strategies> :print b b = [(1 : (_t7::[Integer])),(1 : (_t8::[Integer])), (1 : (_t9::[Integer])),(1 : (_t10::[Integer])), (1 : (_t11::[Integer]))] Prelude Control.Parallel.Strategies> seqList rnf b () Prelude Control.Parallel.Strategies> :print b b = [[1],[1,2],[1,2,3],[1,2,3,4],[1,2,3,4,5]] I'm a newbie, but I believe that a typical usage would be something like: let myList = someExpression `using` seqList r0 where 'using' ensures traversal of the expression's spine. Graham