On Wed, 26 Jan 2005, Simon Marlow wrote:
When using the ordinary 2-generation collector, memory in use will tend to be 2-3 times larger than the actual residency, because this is a copying collector.
Ok. Perhaps you want a different name for this item in the reports?
On the other hand, I found 10mb, 5mb, and 2.5mb maximum residency, but that is still ~100 bytes per record.
But 100 bytes per record in a (Map Int Int) seems awefully high. A 3 record Map would be: (Bin 3 2 2 (Bin 1 1 1 Tip Tip) (Bin 1 3 3 Tip Tip)) Here is how I account for memory: size qty subtotal Ints 4 9 36 Pointers 4 7 28 Constructors 4 7 28 ----------------------------------- Total 92 Num Pairs 3 Bytes/Pair 31 Is this accounting correct? Where do the other 70 bytes/item go?
Lastly, I tried "example +RTS -p -K5M -hc" and then looked at the resulting graph (attachment #2) and it shows a total of 1.6Mb heap allocated and if that data is correct, it does correspond roughly to our 20-30 bytes per record estimate.
That profile doesn't include the stack, which sounds reasonable.
So I guess all the extra memory is actually being allocated on the stack rather than the heap which is a sign of a spaceleak. If I solve the spaceleak I get the 70 bytes per item back.
I haven't looked at your code in detail, hopefully someone else can shed some light on why you're building up such a large stack in the first place.
The code is just this 6 line program: import qualified Map zipped =zip [1..] [1..100000]::[(Int,Int)] untup f (x,y) = f x y produce = foldr (untup Map.insert) Map.empty zipped fm = length $ Map.keys produce main = print $ fm Can someone tell me what I can do to make this more efficient? -Alex- ______________________________________________________________ S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com