staafmeister
The list you give prod is also 10 MB so it not a terribly inefficient program. It's a factor of 2.
No, it is not. The expression: prod [1..1000] + sum [1..1000] will first evaluate prod [1..1000], generating values from 1 to 1000 lazily and accumulating the product incrementally, allowing the used values to be GC'ed. Then the sum will be evaluated similarly, and the two results will be added. It will run in constant - and very small space. If you do common sub-expression elimination (CSE), and write let ls = [1..1000] in prod ls + sum ls the product will force ls to be evaluated, but since there is another reference to this value (for the sum), it cannot be garbage collected. Thus, the difference is more like a factor of 1000, which is a big deal. -k -- If I haven't seen further, it is by standing in the footprints of giants