use new parser which supports the syntax in GF.Grammar.Grammar directly

This commit is contained in:
krasimir
2009-03-16 14:10:30 +00:00
parent 5597cff5cb
commit a391c69fd3
16 changed files with 1031 additions and 4915 deletions
+2 -10
View File
@@ -1,22 +1,19 @@
module GF.Grammar.API (
Grammar,
emptyGrammar,
pTerm,
ppTerm,
checkTerm,
computeTerm,
showTerm,
TermPrintStyle(..), TermPrintQual(..),
) where
import GF.Source.ParGF
import GF.Source.SourceToGrammar (transExp)
import GF.Grammar.Grammar
import GF.Infra.Ident
import GF.Infra.Modules (greatestResource)
import GF.Compile.GetGrammar
import GF.Grammar.Macros
import GF.Grammar.Parser
import GF.Grammar.Printer
import GF.Grammar.Grammar
import GF.Compile.Rename (renameSourceTerm)
import GF.Compile.CheckGrammar (justCheckLTerm)
@@ -33,11 +30,6 @@ type Grammar = SourceGrammar
emptyGrammar :: Grammar
emptyGrammar = emptySourceGrammar
pTerm :: String -> Err Term
pTerm s = do
e <- pExp $ myLexer (BS.pack s)
transExp e
checkTerm :: Grammar -> Term -> Err Term
checkTerm gr t = do
mo <- maybe (Bad "no source grammar in scope") return $ greatestResource gr