[英]How can I parse GHC core?
我认为最简单的方法是放弃这些外部核心库,直接使用GHC作为库。 基于Haskell wiki中的这个示例 ,我们可以创建一个将模块转换为Core的简单函数:
import GHC
import GHC.Paths (libdir)
import HscTypes (mg_binds)
import CoreSyn
import DynFlags
import Control.Monad ((<=<))
compileToCore :: String -> IO [CoreBind]
compileToCore modName = runGhc (Just libdir) $ do
setSessionDynFlags =<< getSessionDynFlags
target <- guessTarget (modName ++ ".hs") Nothing
setTargets [target]
load LoadAllTargets
ds <- desugarModule <=< typecheckModule <=< parseModule <=< getModSummary $ mkModuleName modName
return $ mg_binds . coreModule $ ds
这是一个愚蠢的示例用法,用于处理其输出以计算案例数:
-- Silly example function that analyzes Core
countCases :: [CoreBind] -> Int
countCases = sum . map countBind
where
countBind (NonRec _ e) = countExpr e
countBind (Rec bs) = sum . map (countExpr . snd) $ bs
countExpr (Case e _ _ alts) = countExpr e + sum (map countAlt alts)
countExpr (App f e) = countExpr f + countExpr e
countExpr (Lam _ e) = countExpr e
countExpr (Let b e) = countBind b + countExpr e
countExpr (Cast e _) = countExpr e
countExpr (Tick _ e) = countExpr e
countExpr _ = 0
countAlt (_, _, rhs) = 1 + countExpr rhs
让我们在你的例子中运行它:
main :: IO ()
main = do
core <- compileToCore "MyNot"
print $ countCases core
按预期输出2。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.