Copyright | (c) Varun Gandhi 2018 |
---|---|
License | BSD-style (see the file LICENSE) |
Maintainer | theindigamer15@gmail.com |
Stability | experimental |
Portability | portable |
Safe Haskell | Safe |
Language | Haskell2010 |
This is a short (?) tutorial describing how you can use the Edit
module
to help you with writing dataflow analysis code for a compiler. The example
is a bit artificial for the sake of relative conciseness -- if you have a
better suggestion, or find any mistakes, please let me know on the Github
issue tracker.
Synopsis
TL;DR
Get a fixed point from applying a sequence of transformations.
import Data.Edit (Edit, edits, polish, (>=>)) mkAwesome1 :: Foo -> Maybe Foo ... mkAwesomeN :: Foo -> Maybe Foo mkAwesomeAny :: Foo -> Edit Foo mkAwesomeAny = foldr (\f acc -> acc >=> (f `edits`)) pure [mkAwesome1, ..., mkAwesomeN] mkAsAwesomeAsPossible :: Foo -> Foo mkAsAwesomeAsPossible = polish mkAwesomeAny
Transform a recursive data structure, keeping track of whether it was changed or not, and feed the result to some high-level dataflow analysis function.
import DataFlowLibrary import PlatedLibrary import Data.Edit (Edit, edits, toMaybe) instance FancyPlate Foo where ... mkAwesome :: Foo -> Maybe Foo mkAwesome = ... mkTotallyAwesome :: Foo -> Edit Foo mkTotallyAwesome = transformM (mkAwesome `edits`) dataFlowAnalysis = dataFlowLibFn (toMaybe . mkTotallyAwesome)
Setup
The examples here use the
Uniplate and
Containers libraries.
If you want to
follow along as we proceed, you will want to supply the package flag
tutorial
and maybe read the docs in your browser.
If you're using cabal
, this can be done using
(tested using cabal-install
2.0.0.1):
cabal new-configure --flags="tutorial" cabal new-build cabal new-haddock
If you're using stack
, the same can be done using:
stack build --flag=edit:tutorial stack haddock --flag=edit:tutorial --open edit
Tutorial
Let's define a toy language L
with Int
s and addition.
newtype Ident = Ident String deriving (Show, Eq) data Expr = Val Int | Var Ident | Add Expr Expr deriving (Show, Eq)
Q. How would you implement constant folding for the Expr
type?
- Write the recursion by hand. While this is easy enough to do since
Expr
only has a few constructors, this isn't very practical when you have lots of constructors. The exact point where you recognize that this is a recursive descent into unmaintainability depends on your personal boilerplate threshold. - Use recursion schemes and get lost in the unfathomable type errors (I'm half-joking). While this is a reasonable approach, we're not going to follow this here.
- Use a generics library. For simplicity, we'll be using Uniplate here.
The particular functions that are relevant at the moment are
rewrite
andtransform
. Let's userewrite
.
{-# LANGUAGE DeriveDataTypeable #-} import Data.Data import Data.Generics.Uniplate.Data newtype Ident = Ident String deriving (Show, Eq, Typeable, Data) data Expr = Val Int | Var Ident | Add Expr Expr deriving (Show, Eq, Typeable, Data) constFold :: Expr -> Expr constFold e = rewrite go e where go (Add (Val i) (Val j)) = Just (Val (i + j)) go _ = Nothing
Test that the implementation works as expected.
>>>
two = Add (Val 1) (Val 1)
>>>
four = Add (Val 2) (Val 2)
>>>
constFold (Add two four)
Val 6>>>
constFold (Add (Var "x") two)
Add (Var "x") (Val 2)
Let's say we add assignment statements to the language and write a function
to do constant propagation. First we add a substitute
function.
import Data.Map (Map) import qualified Data.Map as Map newtype Ident = Ident String deriving (Eq, Ord, Show, Typeable, Data) substitute :: Map Ident Int -> Expr -> Expr substitute m e = rewrite go e where go (Var x) = Val <$> Map.lookup x m go _ = Nothing
Let's test this out.
>>>
x = Var (Ident "x")
>>>
quadrupleX = Add x (Add x (Add x x))
>>>
m1 = Map.fromList [(Ident "x", 5)]
>>>
substitute m1 quadrupleX
Add (Val 5) (Add (Val 5) (Add (Val 5) (Val 5)))
Finally putting all the pieces together. We can use the polish
function
to find the fixed point, which (in this case) is a fancy way of saying that
we keep iterating until we have a Clean
(unchanged) value.
constFoldAndPropPass :: [Stmt] -> [Stmt] constFoldAndPropPass = polish (constFoldPass >=> constPropPass)
We're not done yet though! We still need to check that this works :P.
>>>
[w, x, y] = map Ident ["w", "x", "y"]
>>>
s1 = w := Add (Val 1) (Val 2)
>>>
s2 = x := Add (Var w) (Var w)
>>>
s3 = y := Add (Var w) (Add (Val 1) (Var x))
>>>
s4 = x := Add (Var y) (Var y)
>>>
s5 = y := Add (Var w) (Var x)
>>>
constFoldAndPropPass [s1, s2, s3, s4, s5]
[Ident "w" := Val 3,Ident "x" := Val 6,Ident "y" := Val 10,Ident "x" := Val 20,Ident "y" := Val 23]
Yup, it works! For fun, let's see the transformation process in action.
We can do this using the iterations
function.
>>>
pprint = putStr . unlines . map (unlines . map show)
>>>
pprint $ iterations (constFoldPass >=> constPropPass) [s1, s2, s3, s4, s5]
The output shows the full history, with the final result that we obtained earlier at the end.
Ident "w" := Add (Val 1) (Val 2) Ident "x" := Add (Var (Ident "w")) (Var (Ident "w")) Ident "y" := Add (Var (Ident "w")) (Add (Val 1) (Var (Ident "x"))) Ident "x" := Add (Var (Ident "y")) (Var (Ident "y")) Ident "y" := Add (Var (Ident "w")) (Var (Ident "x")) Ident "w" := Val 3 Ident "x" := Add (Val 3) (Val 3) Ident "y" := Add (Val 3) (Add (Val 1) (Var (Ident "x"))) Ident "x" := Add (Var (Ident "y")) (Var (Ident "y")) Ident "y" := Add (Val 3) (Var (Ident "x")) Ident "w" := Val 3 Ident "x" := Val 6 Ident "y" := Add (Val 3) (Add (Val 1) (Val 6)) Ident "x" := Add (Var (Ident "y")) (Var (Ident "y")) Ident "y" := Add (Val 3) (Var (Ident "x")) Ident "w" := Val 3 Ident "x" := Val 6 Ident "y" := Val 10 Ident "x" := Add (Val 10) (Val 10) Ident "y" := Add (Val 3) (Var (Ident "x")) Ident "w" := Val 3 Ident "x" := Val 6 Ident "y" := Val 10 Ident "x" := Val 20 Ident "y" := Add (Val 3) (Val 20) Ident "w" := Val 3 Ident "x" := Val 6 Ident "y" := Val 10 Ident "x" := Val 20 Ident "y" := Val 23
Fin.