108 lines
2.4 KiB
C
108 lines
2.4 KiB
C
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/*
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* (c) copyright 1987 by the Vrije Universiteit, Amsterdam, The Netherlands.
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* See the copyright notice in the ACK home directory, in the file "Copyright".
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*/
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/* $Header$ */
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/* Lint evaluation order checking */
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#include "lint.h"
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#ifdef LINT
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#include <alloc.h> /* for st_free */
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#include "interface.h"
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#include "assert.h"
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#include "arith.h" /* definition arith */
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#include "label.h" /* definition label */
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#include "expr.h"
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#include "idf.h"
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#include "def.h"
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#include "code.h" /* RVAL etc */
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#include "LLlex.h"
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#include "Lpars.h"
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#include "stack.h"
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#include "type.h"
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#include "level.h"
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#include "nofloat.h"
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#include "l_lint.h"
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#include "l_state.h"
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extern char *symbol2str();
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PRIVATE check_ev_order();
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check_and_merge(expr, espp, esp)
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struct expr *expr;
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struct expr_state **espp, *esp;
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{
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/* Checks for undefined evaluation orders in case of a non-sequencing operator.
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* In addition the sets of used and set variables of both expressions are
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* united.
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* *espp will be pointing to this new list. esp is used for this list.
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*/
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register struct expr_state **pp, *p1, *p2;
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int oper = expr->OP_OPER;
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int is_sequencer =
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(oper == '?' || oper == OR || oper == AND || oper ==',');
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for (p1 = *espp; p1; p1 = p1->next) {
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/* scan the list esp for the same variable */
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p2 = esp;
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pp = &esp;
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while (p2) {
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if ( /* p1 and p2 refer to the same location */
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p1->es_idf == p2->es_idf
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&& p1->es_offset == p2->es_offset
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) {
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/* check */
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if (!is_sequencer)
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check_ev_order(p1, p2, expr);
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/* merge the info */
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p1->es_used |= p2->es_used;
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p1->es_referred |= p2->es_referred;
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p1->es_set |= p2->es_set;
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/* and remove the entry from esp */
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*pp = p2->next;
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free_expr_state(p2);
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p2 = *pp;
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}
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else {
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/* skip over the entry in esp */
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pp = &p2->next;
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p2 = p2->next;
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}
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}
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}
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/* If there is anything left in the list esp, this is put in
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front of the list *espp is now pointing to, and *espp will be
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left pointing to this new list.
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*/
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if (!esp)
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return;
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p1 = *espp;
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*espp = esp;
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while (esp->next)
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esp = esp->next;
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esp->next = p1;
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}
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PRIVATE
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check_ev_order(esp1, esp2, expr)
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struct expr_state *esp1, *esp2;
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struct expr *expr;
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{
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if ( (esp1->es_used && esp2->es_set)
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|| (esp1->es_set && esp2->es_used)
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|| (esp1->es_set && esp2->es_set)
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) {
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expr_warning(expr,
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"result of %s depends on evaluation order on %s",
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symbol2str(expr->OP_OPER),
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esp1->es_idf->id_text);
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}
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}
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#endif LINT
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