Changeset 7943

Timestamp:

11/13/05 04:54:04 (3 years ago)

Author:

luqui

Message:

Fixed '$foo'.perl (wasn't properly escaping the $).
In the process, I blessed and implemented "\d[123]".

Files:

• src/Pugs/Lexer.hs (modified) (2 diffs)
• src/Pugs/Pretty.hs (modified) (1 diff)
• t/builtins/perl.t (modified) (1 diff)

src/Pugs/Lexer.hs

@@ -61 +61 @@
 maybeParens :: CharParser st a -> CharParser st a
 maybeParens p = choice [ parens p, p ]
+maybeBrackets :: CharParser st a -> CharParser st a
+maybeBrackets p = choice [ brackets p, p ]
 
 parens     :: CharParser st a -> CharParser st a
@@ -204 +206 @@
 charNum :: GenParser Char st Char                    
 charNum         = do{ code <- decimal 
-                              <|> do{ char 'o'; number 8 octDigit }
-                              <|> do{ char 'x'; number 16 hexDigit }
-                              <|> do{ char 'd'; number 10 digit }
+                              <|> do{ char 'o'; maybeBrackets $ number 8 octDigit }
+                              <|> do{ char 'x'; maybeBrackets $ number 16 hexDigit }
+                              <|> do{ char 'd'; maybeBrackets $ number 10 digit }
                     ; return (toEnum (fromInteger code))
                     }

src/Pugs/Pretty.hs

@@ -187 +187 @@
 quoted '\r' = "\\r"
 quoted '\n' = "\\n"
-quoted x | x < ' ' = "{chr " ++ show (ord x) ++ "}"
--- XXX is there a more elegant way? Remember that
---   my $str = chr(12) ~ "3"; $str.perl
--- may not perlify to "\123"...
+quoted '$'  = "\\$"
+quoted '@'  = "\\@"
+quoted '%'  = "\\%"
+quoted '&'  = "\\&"
+quoted x | x < ' ' || x > '~' = "\\d[" ++ show (ord x) ++ "]"
 quoted x = [x]
 

t/builtins/perl.t

@@ -9 +9 @@
     # Basic scalar values
     42, 42/10, 4.2, sqrt(2), 3e5, Inf, -Inf, NaN,
-    "a string", "", "\0", "\t", "\n", "\r\n", "\7", '{', '}', "\123",
+    "a string", "", "\0", "\t", "\n", "\r\n", "\7", '{', '}', "\123", '$a @string %with &sigils()',
     ?1, ?0,
     undef,