ack/util/ego/cs/cs_profit.c

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/* $Id$ */
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/*
* (c) copyright 1987 by the Vrije Universiteit, Amsterdam, The Netherlands.
* See the copyright notice in the ACK home directory, in the file "Copyright".
*/
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#include <stdio.h>
#include <string.h>
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#include <em_mnem.h>
#include <em_spec.h>
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#include "../share/types.h"
#include "../share/debug.h"
#include "../share/global.h"
#include "../share/aux.h"
#include "../share/cset.h"
#include "../share/lset.h"
#include "cs.h"
#include "cs_alloc.h"
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#include "cs_aux.h"
#include "cs_debug.h"
#include "cs_avail.h"
#include "cs_partit.h"
STATIC cset addr_modes;
STATIC cset cheaps;
STATIC cset forbidden;
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STATIC cset sli_counts;
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STATIC short LX_threshold;
STATIC short AR_limit;
STATIC bool RM_to_DV;
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STATIC void get_instrs(FILE *f, cset *s_p)
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{
/* Read a set of integers from inputfile f into *s_p.
* Such a set must be delimited by a negative number.
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*/
int instr;
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fscanf(f, "%d", &instr);
while (instr >= 0) {
Cadd((Celem_t) instr, s_p);
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fscanf(f, "%d", &instr);
}
}
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STATIC void choose_cset(FILE *f, cset *s_p, int max)
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{
/* Read two compact sets of integers from inputfile f.
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* Choose the first if we optimize with respect to time,
* the second if we optimize with respect to space, as
* indicated by time_space_ratio.
*/
cset cs1, cs2; /* Two dummy sets. */
*s_p = Cempty_set((short) max);
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cs1 = Cempty_set((short) max);
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get_instrs(f, &cs1);
cs2 = Cempty_set((short) max);
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get_instrs(f, &cs2);
Ccopy_set(time_space_ratio >= 50 ? cs1 : cs2, s_p);
Cdeleteset(cs1); Cdeleteset(cs2);
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}
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void cs_machinit(void *vp)
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{
FILE *f = vp;
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char s[100];
int time, space;
/* Find piece that is relevant for this phase. */
do {
while (getc(f) != '\n');
fscanf(f, "%99s", s);
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} while (strcmp(s, "%%CS"));
/* Choose a set of instructions which must only be eliminated
* if they are at the root of another expression.
*/
choose_cset(f, &addr_modes, sp_lmnem);
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/* Choose a set of cheap instructions; i.e. instructions that
* are cheaper than a move to save the result of such an
* instruction.
*/
choose_cset(f, &cheaps, sp_lmnem);
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/* Read how many lexical levels back an LXL/LXA instruction
* must at least look before it will be eliminated.
*/
fscanf(f, "%d %d", &time, &space);
LX_threshold = time_space_ratio >= 50 ? time : space;
/* Read what the size of an array-element may be,
* before we think that it is to big to replace
* a LAR/SAR of it by AAR LOI/STI <size>.
*/
fscanf(f, "%d", &space);
AR_limit = space;
/* Read whether to convert a remainder RMI/RMU to a division
* DVI/DVU using the formula a % b = a - b * (a / b).
*/
fscanf(f, "%d %d", &time, &space);
RM_to_DV = time_space_ratio >= 50 ? time : space;
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/* Read for what counts we must not eliminate an SLI instruction
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* when it is part of an array-index computation.
*/
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choose_cset(f, &sli_counts, 8 * ws);
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/* Read a set of instructions which we do not want to eliminate.
* Note: only instructions need be given that may in principle
* be eliminated, but for which better code can be generated
* when they stay, and with which is not dealt in the common
* decision routines.
*/
choose_cset(f, &forbidden, sp_lmnem);
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}
Check AAR earlier to prevent LOI/STI unknown size. In ego, the CS phase may convert a LAR/SAR to AAR LOI/STI so it can optimize multiple occurrences of AAR of the same array element. This conversion should not happen if it would LOI/STI a large or unknown size. cs_profit.c okay_lines() checked the size of each occurrence of AAR except the first. If the first AAR was the implicit AAR in a LAR/SAR, then the conversion happened without checking the size. For unknown size, this made a bad LOI -1 or STI -1. Fix by checking the size earlier: if a LAR/SAR has a bad size, then don't enter it as an AAR. This Modula-2 code showed the bug. Given M.def: DEFINITION MODULE M; TYPE S = SET OF [0..95]; PROCEDURE F(a: ARRAY OF S; i, j: INTEGER); END M. and M.mod: (*$R-*) IMPLEMENTATION MODULE M; FROM SYSTEM IMPORT ADDRESS, ADR; PROCEDURE G(s: S; p, q: ADDRESS; t: S); BEGIN s := s; p := p; q := q; t := t; END G; PROCEDURE F(a: ARRAY OF S; i, j: INTEGER); BEGIN G(a[i + j], ADR(a[i + j]), ADR(a[i + j]), a[i + j]) END F; END M. then the bug caused an error: $ ack -mlinuxppc -O3 -c.e M.mod /tmp/Ack_b357d.g, line 57: Argument range error The bug had put LOI -1 in the code, then em_decode got an error because -1 is out of range for LOI. Procedure F has 4 occurrences of `a[i + j]`. The size of `a[i + j]` is 96 bits, or 12 bytes, but the EM code hides the size in an array descriptor, so the size is unknown to CS. The pragma `(*$R-*)` disables a range check on `i + j` so CS can work. EM uses AAR for the 2 `ADR(a[i + j])` and LAR for the other 2 `a[i + j]`. EM pushes the arguments to G in reverse order, so the last `a[i + j]` in Modula-2 is the first LAR in EM. CS found 4 occurrences of AAR. The first AAR was an implicit AAR in LAR. Because of the bug, CS converted this LAR 4 to AAR 4 LOI -1.
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bool may_become_aar(avail_p avp)
{
/* Check whether it is desirable to treat a LAR or SAR as an
* AAR LOI/STI. This depends on the size of the array-elements.
*/
offset sz;
sz = array_elemsize(avp->av_othird);
if (sz == UNKNOWN_SIZE)
return FALSE;
if (time_space_ratio < 50)
return sz <= AR_limit;
return TRUE;
}
bool may_become_dv(void)
{
return RM_to_DV;
}
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STATIC bool sli_no_eliminate(line_p lnp)
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{
/* Return whether the SLI-instruction in lnp is part of
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* an array-index computation, and should not be eliminated.
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*/
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offset cst;
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return lnp->l_prev != (line_p) 0 && INSTR(lnp->l_prev) == op_loc &&
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lnp->l_next != (line_p) 0 && INSTR(lnp->l_next) == op_ads &&
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((cst = off_set(lnp->l_prev)), cst == (Celem_t) cst) &&
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Cis_elem((Celem_t) cst, sli_counts)
;
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}
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STATIC bool gains(avail_p avp)
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{
/* Return whether we can gain something, when we eliminate
* an expression such as in avp. We just glue together some
* heuristics with some user-supplied stuff.
*/
if (Cis_elem(avp->av_instr & BMASK, forbidden))
return FALSE;
if (avp->av_instr == (byte) op_lxa || avp->av_instr == (byte) op_lxl)
return off_set(avp->av_found) >= LX_threshold;
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if (avp->av_instr == (byte) op_sli || avp->av_instr == (byte) op_slu)
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return ! sli_no_eliminate(avp->av_found);
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if (avp->av_instr == (byte) op_ads &&
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avp->av_found->l_prev &&
( INSTR(avp->av_found->l_prev) == op_sli ||
INSTR(avp->av_found->l_prev) == op_slu))
return ! sli_no_eliminate(avp->av_found->l_prev);
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if (Cis_elem(avp->av_instr & BMASK, addr_modes))
return instrgroup(avp->av_found->l_prev) != SIMPLE_LOAD;
if (Cis_elem(avp->av_instr & BMASK, cheaps))
return avp->av_saveloc != (entity_p) 0;
return TRUE;
}
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STATIC bool okay_lines(avail_p avp, occur_p ocp)
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{
Check AAR earlier to prevent LOI/STI unknown size. In ego, the CS phase may convert a LAR/SAR to AAR LOI/STI so it can optimize multiple occurrences of AAR of the same array element. This conversion should not happen if it would LOI/STI a large or unknown size. cs_profit.c okay_lines() checked the size of each occurrence of AAR except the first. If the first AAR was the implicit AAR in a LAR/SAR, then the conversion happened without checking the size. For unknown size, this made a bad LOI -1 or STI -1. Fix by checking the size earlier: if a LAR/SAR has a bad size, then don't enter it as an AAR. This Modula-2 code showed the bug. Given M.def: DEFINITION MODULE M; TYPE S = SET OF [0..95]; PROCEDURE F(a: ARRAY OF S; i, j: INTEGER); END M. and M.mod: (*$R-*) IMPLEMENTATION MODULE M; FROM SYSTEM IMPORT ADDRESS, ADR; PROCEDURE G(s: S; p, q: ADDRESS; t: S); BEGIN s := s; p := p; q := q; t := t; END G; PROCEDURE F(a: ARRAY OF S; i, j: INTEGER); BEGIN G(a[i + j], ADR(a[i + j]), ADR(a[i + j]), a[i + j]) END F; END M. then the bug caused an error: $ ack -mlinuxppc -O3 -c.e M.mod /tmp/Ack_b357d.g, line 57: Argument range error The bug had put LOI -1 in the code, then em_decode got an error because -1 is out of range for LOI. Procedure F has 4 occurrences of `a[i + j]`. The size of `a[i + j]` is 96 bits, or 12 bytes, but the EM code hides the size in an array descriptor, so the size is unknown to CS. The pragma `(*$R-*)` disables a range check on `i + j` so CS can work. EM uses AAR for the 2 `ADR(a[i + j])` and LAR for the other 2 `a[i + j]`. EM pushes the arguments to G in reverse order, so the last `a[i + j]` in Modula-2 is the first LAR in EM. CS found 4 occurrences of AAR. The first AAR was an implicit AAR in LAR. Because of the bug, CS converted this LAR 4 to AAR 4 LOI -1.
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/* Check whether all lines in this occurrence can in
* principle be eliminated; no stores, messages, calls etc.
*/
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register line_p lnp, next;
for (lnp = ocp->oc_lfirst; lnp != (line_p) 0; lnp = next) {
next = lnp != ocp->oc_llast ? lnp->l_next : (line_p) 0;
if (INSTR(lnp) < sp_fmnem || INSTR(lnp) > sp_lmnem)
return FALSE;
if (!stack_group(INSTR(lnp))) {
/* Check for SAR-instruction. */
if (INSTR(lnp) != op_sar || next != (line_p) 0)
return FALSE;
}
}
return TRUE;
}
bool desirable(avail_p avp)
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{
register Lindex i, next;
if (!gains(avp)) {
OUTTRACE("no gain", 0);
SHOWAVAIL(avp);
return FALSE;
}
/* Walk through the occurrences to see whether it is okay to
* eliminate them. If not, remove them from the set.
*/
for (i = Lfirst(avp->av_occurs); i != (Lindex) 0; i = next) {
next = Lnext(i, avp->av_occurs);
if (!okay_lines(avp, occ_elem(i))) {
OUTTRACE("may not eliminate", 0);
# ifdef TRACE
SHOWOCCUR(occ_elem(i));
# endif
oldoccur(occ_elem(i));
Lremove(Lelem(i), &avp->av_occurs);
}
}
return Lnrelems(avp->av_occurs) > 0;
}