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     х€НЦ„dґзМїФ ?ч     х// SPDX-License-Identifier: GPL-2.0
/*
 * Copyright (C) 2002 Roman Zippel <zippel@linux-m68k.org>
 */

#include <ctype.h>
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <xalloc.h>
#include "lkc.h"

#define DEBUG_EXPR	0

static struct expr *expr_eliminate_yn(struct expr *e);

struct expr *expr_alloc_symbol(struct symbol *sym)
{
	struct expr *e = xcalloc(1, sizeof(*e));
	e->type = E_SYMBOL;
	e->left.sym = sym;
	return e;
}

struct expr *expr_alloc_one(enum expr_type type, struct expr *ce)
{
	struct expr *e = xcalloc(1, sizeof(*e));
	e->type = type;
	e->left.expr = ce;
	return e;
}

struct expr *expr_alloc_two(enum expr_type type, struct expr *e1, struct expr *e2)
{
	struct expr *e = xcalloc(1, sizeof(*e));
	e->type = type;
	e->left.expr = e1;
	e->right.expr = e2;
	return e;
}

struct expr *expr_alloc_comp(enum expr_type type, struct symbol *s1, struct symbol *s2)
{
	struct expr *e = xcalloc(1, sizeof(*e));
	e->type = type;
	e->left.sym = s1;
	e->right.sym = s2;
	return e;
}

struct expr *expr_alloc_and(struct expr *e1, struct expr *e2)
{
	if (!e1)
		return e2;
	return e2 ? expr_alloc_two(E_AND, e1, e2) : e1;
}

struct expr *expr_alloc_or(struct expr *e1, struct expr *e2)
{
	if (!e1)
		return e2;
	return e2 ? expr_alloc_two(E_OR, e1, e2) : e1;
}

struct expr *expr_copy(const struct expr *org)
{
	struct expr *e;

	if (!org)
		return NULL;

	e = xmalloc(sizeof(*org));
	memcpy(e, org, sizeof(*org));
	switch (org->type) {
	case E_SYMBOL:
		e->left = org->left;
		break;
	case E_NOT:
		e->left.expr = expr_copy(org->left.expr);
		break;
	case E_EQUAL:
	case E_GEQ:
	case E_GTH:
	case E_LEQ:
	case E_LTH:
	case E_UNEQUAL:
		e->left.sym = org->left.sym;
		e->right.sym = org->right.sym;
		break;
	case E_AND:
	case E_OR:
	case E_LIST:
		e->left.expr = expr_copy(org->left.expr);
		e->right.expr = expr_copy(org->right.expr);
		break;
	default:
		fprintf(stderr, "can't copy type %d\n", e->type);
		free(e);
		e = NULL;
		break;
	}

	return e;
}

void expr_free(struct expr *e)
{
	if (!e)
		return;

	switch (e->type) {
	case E_SYMBOL:
		break;
	case E_NOT:
		expr_free(e->left.expr);
		break;
	case E_EQUAL:
	case E_GEQ:
	case E_GTH:
	case E_LEQ:
	case E_LTH:
	case E_UNEQUAL:
		break;
	case E_OR:
	case E_AND:
		expr_free(e->left.expr);
		expr_free(e->right.expr);
		break;
	default:
		fprintf(stderr, "how to free type %d?\n", e->type);
		break;
	}
	free(e);
}

static int trans_count;

#define e1 (*ep1)
#define e2 (*ep2)

/*
 * expr_eliminate_eq() helper.
 *
 * Walks the two expression trees given in 'ep1' and 'ep2'. Any node that does
 * not have type 'type' (E_OR/E_AND) is considered a leaf, and is compared
 * against all other leaves. Two equal leaves are both replaced with either 'y'
 * or 'n' as appropriate for 'type', to be eliminated later.
 */
static void __expr_eliminate_eq(enum expr_type type, struct expr **ep1, struct expr **ep2)
{
	/* Recurse down to leaves */

	if (e1->type == type) {
		__expr_eliminate_eq(type, &e1->left.expr, &e2);
		__expr_eliminate_eq(type, &e1->right.expr, &e2);
		return;
	}
	if (e2->type == type) {
		__expr_eliminate_eq(type, &e1, &e2->left.expr);
		__expr_eliminate_eq(type, &e1, &e2->right.expr);
		return;
	}

	/* e1 and e2 are leaves. Compare them. */

	if (e1->type == E_SYMBOL && e2->type == E_SYMBOL &&
	    e1->left.sym == e2->left.sym &&
	    (e1->left.sym == &symbol_yes || e1->left.sym == &symbol_no))
		return;
	if (!expr_eq(e1, e2))
		return;

	/* e1 and e2 are equal leaves. Prepare them for elimination. */

	trans_count++;
	expr_free(e1); expr_free(e2);
	switch (type) {
	case E_OR:
		e1 = expr_alloc_symbol(&symbol_no);
		e2 = expr_alloc_symbol(&symbol_no);
		break;
	case E_AND:
		e1 = expr_alloc_symbol(&symbol_yes);
		e2 = expr_alloc_symbol(&symbol_yes);
		break;
	default:
		;
	}
}

/*
 * Rewrites the expressions 'ep1' and 'ep2' to remove operands common to both.
 * Example reductions:
 *
 *	ep1: A && B           ->  ep1: y
 *	ep2: A && B && C      ->  ep2: C
 *
 *	ep1: A || B           ->  ep1: n
 *	ep2: A || B || C      ->  ep2: