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407 lines
7.9 KiB
Plaintext
407 lines
7.9 KiB
Plaintext
%{
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/* Expression parsing for plural form selection.
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Copyright (C) 2000, 2001 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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Written by Ulrich Drepper <drepper@cygnus.com>, 2000.
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The GNU C Library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the License, or (at your option) any later version.
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The GNU C Library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with the GNU C Library; if not, write to the Free
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Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
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02111-1307 USA. */
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#ifdef HAVE_CONFIG_H
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# include <config.h>
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#endif
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#include <stdlib.h>
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#include "gettextP.h"
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/* Names for the libintl functions are a problem. They must not clash
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with existing names and they should follow ANSI C. But this source
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code is also used in GNU C Library where the names have a __
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prefix. So we have to make a difference here. */
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#ifdef _LIBC
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# define FREE_EXPRESSION __gettext_free_exp
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#else
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# define FREE_EXPRESSION gettext_free_exp__
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# define __gettextparse gettextparse__
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#endif
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#define YYLEX_PARAM &((struct parse_args *) arg)->cp
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#define YYPARSE_PARAM arg
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%}
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%pure_parser
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%expect 7
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%union {
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unsigned long int num;
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enum operator op;
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struct expression *exp;
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}
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%{
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/* Prototypes for local functions. */
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static struct expression *new_exp PARAMS ((int nargs, enum operator op,
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struct expression * const *args));
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static inline struct expression *new_exp_0 PARAMS ((enum operator op));
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static inline struct expression *new_exp_1 PARAMS ((enum operator op,
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struct expression *right));
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static struct expression *new_exp_2 PARAMS ((enum operator op,
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struct expression *left,
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struct expression *right));
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static inline struct expression *new_exp_3 PARAMS ((enum operator op,
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struct expression *bexp,
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struct expression *tbranch,
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struct expression *fbranch));
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static int yylex PARAMS ((YYSTYPE *lval, const char **pexp));
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static void yyerror PARAMS ((const char *str));
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/* Allocation of expressions. */
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static struct expression *
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new_exp (nargs, op, args)
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int nargs;
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enum operator op;
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struct expression * const *args;
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{
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int i;
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struct expression *newp;
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/* If any of the argument could not be malloc'ed, just return NULL. */
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for (i = nargs - 1; i >= 0; i--)
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if (args[i] == NULL)
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goto fail;
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/* Allocate a new expression. */
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newp = (struct expression *) malloc (sizeof (*newp));
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if (newp != NULL)
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{
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newp->nargs = nargs;
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newp->operation = op;
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for (i = nargs - 1; i >= 0; i--)
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newp->val.args[i] = args[i];
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return newp;
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}
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fail:
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for (i = nargs - 1; i >= 0; i--)
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FREE_EXPRESSION (args[i]);
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return NULL;
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}
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static inline struct expression *
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new_exp_0 (op)
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enum operator op;
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{
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return new_exp (0, op, NULL);
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}
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static inline struct expression *
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new_exp_1 (op, right)
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enum operator op;
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struct expression *right;
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{
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struct expression *args[1];
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args[0] = right;
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return new_exp (1, op, args);
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}
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static struct expression *
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new_exp_2 (op, left, right)
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enum operator op;
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struct expression *left;
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struct expression *right;
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{
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struct expression *args[2];
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args[0] = left;
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args[1] = right;
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return new_exp (2, op, args);
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}
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static inline struct expression *
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new_exp_3 (op, bexp, tbranch, fbranch)
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enum operator op;
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struct expression *bexp;
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struct expression *tbranch;
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struct expression *fbranch;
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{
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struct expression *args[3];
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args[0] = bexp;
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args[1] = tbranch;
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args[2] = fbranch;
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return new_exp (3, op, args);
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}
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%}
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/* This declares that all operators have the same associativity and the
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precedence order as in C. See [Harbison, Steele: C, A Reference Manual].
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There is no unary minus and no bitwise operators.
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Operators with the same syntactic behaviour have been merged into a single
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token, to save space in the array generated by bison. */
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%right '?' /* ? */
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%left '|' /* || */
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%left '&' /* && */
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%left EQUOP2 /* == != */
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%left CMPOP2 /* < > <= >= */
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%left ADDOP2 /* + - */
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%left MULOP2 /* * / % */
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%right '!' /* ! */
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%token <op> EQUOP2 CMPOP2 ADDOP2 MULOP2
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%token <num> NUMBER
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%type <exp> exp
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%%
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start: exp
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{
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if ($1 == NULL)
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YYABORT;
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((struct parse_args *) arg)->res = $1;
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}
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;
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exp: exp '?' exp ':' exp
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{
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$$ = new_exp_3 (qmop, $1, $3, $5);
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}
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| exp '|' exp
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{
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$$ = new_exp_2 (lor, $1, $3);
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}
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| exp '&' exp
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{
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$$ = new_exp_2 (land, $1, $3);
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}
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| exp EQUOP2 exp
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{
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$$ = new_exp_2 ($2, $1, $3);
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}
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| exp CMPOP2 exp
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{
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$$ = new_exp_2 ($2, $1, $3);
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}
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| exp ADDOP2 exp
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{
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$$ = new_exp_2 ($2, $1, $3);
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}
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| exp MULOP2 exp
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{
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$$ = new_exp_2 ($2, $1, $3);
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}
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| '!' exp
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{
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$$ = new_exp_1 (lnot, $2);
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}
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| 'n'
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{
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$$ = new_exp_0 (var);
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}
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| NUMBER
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{
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if (($$ = new_exp_0 (num)) != NULL)
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$$->val.num = $1;
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}
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| '(' exp ')'
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{
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$$ = $2;
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}
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;
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%%
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void
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internal_function
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FREE_EXPRESSION (exp)
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struct expression *exp;
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{
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if (exp == NULL)
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return;
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/* Handle the recursive case. */
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switch (exp->nargs)
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{
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case 3:
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FREE_EXPRESSION (exp->val.args[2]);
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/* FALLTHROUGH */
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case 2:
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FREE_EXPRESSION (exp->val.args[1]);
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/* FALLTHROUGH */
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case 1:
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FREE_EXPRESSION (exp->val.args[0]);
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/* FALLTHROUGH */
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default:
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break;
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}
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free (exp);
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}
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static int
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yylex (lval, pexp)
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YYSTYPE *lval;
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const char **pexp;
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{
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const char *exp = *pexp;
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int result;
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while (1)
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{
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if (exp[0] == '\0')
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{
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*pexp = exp;
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return YYEOF;
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}
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if (exp[0] != ' ' && exp[0] != '\t')
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break;
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++exp;
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}
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result = *exp++;
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switch (result)
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{
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case '0': case '1': case '2': case '3': case '4':
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case '5': case '6': case '7': case '8': case '9':
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{
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unsigned long int n = result - '0';
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while (exp[0] >= '0' && exp[0] <= '9')
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{
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n *= 10;
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n += exp[0] - '0';
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++exp;
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}
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lval->num = n;
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result = NUMBER;
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}
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break;
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case '=':
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if (exp[0] == '=')
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{
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++exp;
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lval->op = equal;
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result = EQUOP2;
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}
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else
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result = YYERRCODE;
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break;
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case '!':
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if (exp[0] == '=')
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{
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++exp;
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lval->op = not_equal;
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result = EQUOP2;
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}
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break;
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case '&':
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case '|':
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if (exp[0] == result)
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++exp;
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else
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result = YYERRCODE;
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break;
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case '<':
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if (exp[0] == '=')
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{
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++exp;
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lval->op = less_or_equal;
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}
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else
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lval->op = less_than;
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result = CMPOP2;
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break;
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case '>':
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if (exp[0] == '=')
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{
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++exp;
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lval->op = greater_or_equal;
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}
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else
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lval->op = greater_than;
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result = CMPOP2;
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break;
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case '*':
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lval->op = mult;
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result = MULOP2;
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break;
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case '/':
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lval->op = divide;
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result = MULOP2;
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break;
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case '%':
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lval->op = module;
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result = MULOP2;
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break;
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case '+':
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lval->op = plus;
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result = ADDOP2;
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break;
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case '-':
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lval->op = minus;
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result = ADDOP2;
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break;
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case 'n':
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case '?':
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case ':':
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case '(':
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case ')':
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/* Nothing, just return the character. */
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break;
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case ';':
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case '\n':
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case '\0':
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/* Be safe and let the user call this function again. */
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--exp;
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result = YYEOF;
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break;
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default:
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result = YYERRCODE;
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#if YYDEBUG != 0
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--exp;
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#endif
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break;
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}
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*pexp = exp;
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return result;
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}
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static void
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yyerror (str)
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const char *str;
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{
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/* Do nothing. We don't print error messages here. */
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}
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