关于“ parMap”的推理

Question

Given:

Prelude Control.Parallel.Strategies> :t parMap
parMap :: Strategy b -> (a -> b) -> [a] -> [b]

Prelude Control.Parallel.Strategies> :i Strategy
type Strategy a = a -> Eval a
        -- Defined in `Control.Parallel.Strategies'

My understanding then is that parMap can be expanded, replacing b with b -> Eval b to:

parMap :: (a -> b -> Eval b) -> (a -> b -> Eval b) -> [b -> Eval b]

This seems more complicated than the example version in Parallel and Concurrent Haskell :

parMap :: (a -> b) -> [a] -> Eval [b]
parMap f [] = return []
parMap f (a:as) = do
   b <- rpar (f a)
   bs <- parMap f as
   return (b:bs)

In the standard library implementation of parMap , what's the meaning of Strategy b , namely in a -> b -> Eval b ?

Answer 1

There is a mistake in your interpretation, Strategy is not a type class but a type alias. If you check the signature of parMap , Strategy b is followed by -> and not by => which means that parMap expects a Strategy b as first argument. Strategy b is alias for b -> Eval b which means that the signature of parMap can be expanded to:

parMap :: (b -> Eval b) -> (a -> b) -> [a] -> [b]

The book parMap is not the same as the one above because its signature is

parMap :: (a -> b) -> [a] -> Eval [b]

The main difference is that the first version uses the given Strategy b to compute a [b] from the arguments (a -> b) -> [a] while the second version doesn't compute a [b] but a Eval [b] . Eval [b] specifies how to produce a [b] and you must call runEval to use it.

The first version is easier to use in a program because you don't need to call runEval and it's also better because the strategy to evaluate each element of the list is not predefined, like in the book where the strategy is rpar , but it's an argument. This means that the first version is more generic. Have a look at the basic strategies to better understand what you can pass to the standard parMap .