How to turn LHExpr GhcPs into CoreExpr
Yiyun Liu
liuyiyun at terpmail.umd.edu
Tue Jan 21 19:21:38 UTC 2020
Hi ghc-devs,
I've been trying to implementing a function with the signature:
elaborateExpr :: GhcMonad m => String -> m CoreExpr
It should take a string of the form:
"\x y -> x <> y :: Semigroup a => a -> a -> a"
and give back a core expression with the dictionaries filled in:
\ ($dSem :: Semigroup a) (x :: a) (y :: a) -> (<>) $dSem x y
The goal is to use the function to add elaboration support for
liquidhaskell.
I looked into the implementation of exprType
<https://hackage.haskell.org/package/ghc-8.6.5/docs/CoreUtils.html#v:exprType>
and defined my own elaborateExpr
<https://github.com/yiyunliu/ghc-elaboration-test/blob/30b3307469df15789cf1a323bed96ca42e84aeb5/src/Main.hs#L58>
similarly by calling desugarExpr
<https://hackage.haskell.org/package/ghc-8.6.5/docs/Desugar.html#v:deSugarExpr>
on the expression (which has type LHsExpr GhcTcId) returned by tcInferSigma.
GhcTcId is a synonym of GhcTc so the program I wrote typechecks, but
it's definitely not right. The elaborateExpr function I defined would
return something even when the expression doesn't typecheck, or
occasionally give a runtime exception:
ghc-elaboration-test: panic! (the 'impossible' happened)
(GHC version 8.6.5 for x86_64-unknown-linux):
dsEvBinds
I must have broken some invariants somehow.
What is the correct way of defining such a function (takes a string and
returns a CoreExpr)? It appears to me that I should convert LHsExpr
GhcPs into LHsExpr GhcTc first before calling deSugarExpr, but I don't
know how.
Thank you,
- Yiyun
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