ghc-7.4.2: The GHC API

Safe HaskellNone




Vectorisation of expressions.


Vectorise polymorphic expressions with special cases for right-hand sides of particular



:: Bool

When vectorising the RHS of a binding: is that binding a loop breaker?

-> [Var] 
-> CoreExprWithFVs 
-> VM (Inline, Bool, VExpr) 

Vectorise a polymorphic expression.

vectDictExpr :: CoreExpr -> VM CoreExprSource

Vectorise the body of a dfun.

Dictionary computations are special for the following reasons. The application of dictionary functions are always saturated, so there is no need to create closures. Dictionary computations don't depend on array values, so they are always scalar computations whose result we can replicate (instead of executing them in parallel).

NB: To keep things simple, we are not rewriting any of the bindings introduced in a dictionary computation. Consequently, the variable case needs to deal with cases where binders are in the vectoriser environments and where that is not the case.



:: Bool

Was the function marked as scalar by the user?

-> [Var]

Functions names in same recursive binding group

-> CoreExpr

Expression to be vectorised

-> VM VExpr 

Vectorise an expression of functional type, where all arguments and the result are of primitive types (i.e., Int, Float, Double etc., which have instances of the Scalar type class) and which does not contain any subcomputations that involve parallel arrays. Such functionals do not requires the full blown vectorisation transformation; instead, they can be lifted by application of a member of the zipWith family (i.e., map, zipWith, zipWith3', etc.)

Dictionary functions are also scalar functions (as dictionaries themselves are not vectorised, instead they become dictionaries of vectorised methods). We treat them differently, though see Note [Scalar dfuns] in Vectorise.



:: Var

Original dfun

-> [Var]

Functions names in same recursive binding group

-> VM CoreExpr 

Vectorise a dictionary function that has a 'VECTORISE SCALAR instance' pragma.

In other words, all methods in that dictionary are scalar functions — to be vectorised with vectScalarFun. The dictionary function itself may be a constant, though.

NB: You may think that we could implement this function guided by the struture of the Core expression of the right-hand side of the dictionary function. We cannot proceed like this as vectScalarDFun must also work for *imported* dfuns, where we don't necessarily have access to the Core code of the unvectorised dfun.

Here an example — assume,

 class Eq a where { (==) :: a -> a -> Bool }
 instance (Eq a, Eq b) => Eq (a, b) where { (==) = ... }
 {-# VECTORISE SCALAR instance Eq (a, b) }

The unvectorised dfun for the above instance has the following signature:

 $dEqPair :: forall a b. Eq a -> Eq b -> Eq (a, b)

We generate the following (scalar) vectorised dfun (liberally using TH notation):

 $v$dEqPair :: forall a b. V:Eq a -> V:Eq b -> V:Eq (a, b)
 $v$dEqPair = /\a b -> \dEqa :: V:Eq a -> \dEqb :: V:Eq b ->
                D:V:Eq $(vectScalarFun True recFns 
                         [| (==) @(a, b) ($dEqPair @a @b $(unVect dEqa) $(unVect dEqb)) |])

NB: * '(,)' vectorises to '(,)' — hence, the type constructor in the result type remains the same. * We share the '$(unVect di)' sub-expressions between the different selectors, but duplicate the application of the unvectorised dfun, to enable the dictionary selection rules to fire.