edit-1.0.1.0: A monad for rewriting things.

Copyright(c) Varun Gandhi 2018
LicenseBSD-style (see the file LICENSE)
Maintainertheindigamer15@gmail.com
Stabilityexperimental
Portabilityportable
Safe HaskellSafe
LanguageHaskell2010

Data.Edit.Tutorial

Contents

Description

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 Ints 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?

    1. 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.
    2. 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.
    3. Use a generics library. For simplicity, we'll be using Uniplate here. The particular functions that are relevant at the moment are rewrite and transform. Let's use rewrite.
    {-# 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.