I have remembered the name correctly, haven't I?
Would it gain us much implementing De Morgan's theorem in the peephole
optimiser?
nick@Bagpuss [nick]$ perl -le 'for $l (0,1) {for $r (0, 1) {print 0+(!$l && !$r) }}'
1
0
0
0
nick@Bagpuss [nick]$ perl -le 'for $l (0,1) {for $r (0, 1) {print 0+!($l || $r) }}'
1
0
0
0
nick@Bagpuss [nick]$ perl -MO=Terse -e 'print 0+(!$l && !$r)'
LISTOP (0x164048) leave [1]
OP (0x164070) enter
COP (0x164008) nextstate
LISTOP (0x163fc0) print
OP (0x163fe8) pushmark
BINOP (0x163f98) add [1]
SVOP (0x163db0) const IV (0xed2b8) 0
UNOP (0x163f78) null
LOGOP (0x10db20) and
UNOP (0x163e68) not
UNOP (0x163e48) null [15]
SVOP (0x163dd0) gvsv GV (0x10a974) *l
UNOP (0x163f58) not
UNOP (0x163f38) null [15]
SVOP (0x163e88) gvsv GV (0x10a98c) *r
-e syntax OK
nick@Bagpuss [nick]$ perl -MO=Terse -e 'print 0+!($l || $r)'
LISTOP (0x164028) leave [1]
OP (0x164050) enter
COP (0x163fe8) nextstate
LISTOP (0x163fa0) print
OP (0x163fc8) pushmark
BINOP (0x163f78) add [1]
SVOP (0x163db0) const IV (0xed2b8) 0
UNOP (0x163f58) not
UNOP (0x163f38) null
LOGOP (0x10db20) or
UNOP (0x163e48) null [15]
SVOP (0x163dd0) gvsv GV (0x10a974) *l
UNOP (0x163f18) null [15]
SVOP (0x163e68) gvsv GV (0x10a98c) *r
-e syntax OK
For "much" equal to 1 op in total.
I think that the answer is "no, do it by hand if it matters that much",
doesn't it?
This also might be a perl6 question, for a more "serious" -O2 optimiser.
Hmm. Would parrot benefit from nand and nor ops?
[beware of cross posting when replying]
Nicholas Clark
--
EMCFT http://www.ccl4.org/~nick/CV.html
