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Allow labels with same name that a previous typedef

Labels have a separate namespace, so it is correct this combination,
but the problem is the lexer returns a different token for an
identifier after a typedef for it. For example:

	main()
	{
		typedef int pepe;

	pepe:	return 0;
	}

The easier solution is a hack, where we modify the grammar to
accept an IDEN or a TYPE when creating labels.

--- a/cc1/stmt.c

+++ b/cc1/stmt.c

@@ -170,11 +170,16 @@

 static void

 Label(Symbol *lbreak, Symbol *lcont, Caselist *lswitch)

 {

-	emitlabel(label(yytext, 1));

-

-	expect(IDEN);

-	expect(':');

-	stmt(lbreak, lcont, lswitch);

+	switch (yytoken) {

+	case IDEN: case TYPE:

+		emitlabel(label(yytext, 1));

+		next();

+		expect(':');

+		stmt(lbreak, lcont, lswitch);

+		break;

+	default:

+		unexpected();

+	}

 }

 static void

@@ -297,10 +302,15 @@

 		case '}':

 			next();

 			return;

-		case TYPE: case SCLASS: case TQUALIFIER:

+		case TYPE:

+			if (ahead() == ':')

+				goto statement;

+			/* pass through */

+		case SCLASS: case TQUALIFIER:

 			decl();

 			break;

 		default:

+		statement:

 			stmt(lbreak, lcont, lswitch);

 		}

 	}

@@ -325,7 +335,9 @@

 	case CASE:     fun = Case;     break;

 	case DEFAULT:  fun = Default;  break;

 	default:       fun = stmtexp;  break;

-	case IDEN:     fun = (ahead() == ':') ? Label : stmtexp; break;

+	case TYPE: case IDEN:

+		fun = (ahead() == ':') ? Label : stmtexp;

+		break;

 	}

 	(*fun)(lbreak, lcont, lswitch);

 }