Just wanted to explore a few points you made: Jon Cast wrote:
No. A function may not use all of its arguments, so there's no point in evaluating all arguments---you have to find out which ones are actually needed. Terminology note: GHC uses the STG language internally. In STG, evaluation is always triggered by a case expression, which evaluates precisely one expression (not necessarily an argument of anything, either!)
If a function never uses an argument you dont need to evaluate it... remember in dataflow we start from the end of the program working backwards so we know all future demands on function arguments...
Now, (:) cannot `spark' a thread to evaluate xn1 in downFrom, because that would end up sparking infinitely many threads---which may be possible, but nothing you've said suggests that it is.
Again think of lazy lists in their 'stream' interpretation. We model them as a FIFO queue and can set high and low water marks in the FIFO. We can use hysteresis to ensure work gets done in reasonably large chunks. I guess my description didn't include all the details of the model I had in mind... you obviously dont want to unroll all recursive functions... I was thinking along the lines of expressions like let a = c*d + e*f where you can obviously execute the two multiplies in parallel (the multiplies being the arguments to the (+) function. Also it is obvious you don't need a new thread for one of the arguments (a function of one argument would use the same thread)... In the general recursive case: fna x y z = fna (x+1) (y-1) (z*3) again we can do the three simple arithmetic operations in parallel but have to wait for all of them to finish before the call to (fna) can be made. So there isn't an infinite expansion of threads there is one thead that does the recursion and one of the arithmetic operations and two that are forked and reaped each iteration.
This is true for tree reduction, but not for graph reduction. Consider, e.g., the fact that the call to product in choose and the recursive call to downFrom walk down (and hence evaluate) the same list. Since the cons cells making up this list are evaluated by two different functions, some sort of locking is indeed necessary to ensure that you don't end up with two threads for each of the (infinitely many) cons cells in the list.
I suspect I haven't fully understood the difficaulty in doing this, care to enlighten me?
A final point someone made about the cost of starting threads... Surely
The list is read only (this is a declarative language) no locking is necessary on read only data structures. the obvious approach is to start one OS thread per execution unit, and do all the thread starting with the very lightweight haskell threads... I guess in this case I can see where locking would be necessary... Keean