Re^3: Puzzling $| behavior

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Re^4: Puzzling $\| behavior by mwah (Hermit) on Oct 08, 2007 at 10:35 UTC
ikegami: no one has mentioned a system or version where the arguments are evaluated in any order other than left-to-right My answer to the OP was completely wrong, which is what I found out after investigating into the topic. Thats another case where prejudice brings ill-fated conclusions. One simple look into `perl -MO=Bblock thisprog.pl` reveals the left-to-right sequence OP (0x824dc78) enter COP (0x81f30f0) nextstate SVOP (0x824daa0) const [6] IV (0x8167cdc) 1 PADOP (0x8193748) gvsv GV (0x816887c) \| BINOP (0x8189228) sassign COP (0x824c460) nextstate OP (0x824ffa8) pushmark SVOP (0x81892b8) const [7] PV (0x8168804) "first=" PADOP (0x818c5c0) gvsv GV (0x816887c) \| SVOP (0x818dc50) const [8] PV (0x8168810) " second=" PADOP (0x824fdb8) gvsv GV (0x816887c) *\| UNOP (0x824fc80) postinc [4] SVOP (0x824dc58) const [9] PV (0x81688a0) "\n" LISTOP (0x824ff80) print LISTOP (0x824d8e0) leave [1] [download] int the perl. Sorry, I was mistaken by the way one "programs" Perl in C via its interfaces. Thanks to all people who helped clearing this up. I will eventually make an addendum to my first post. Regards mwa	[reply] [d/l] [select]
Re^4: Puzzling $\| behavior by oha (Friar) on Oct 08, 2007 at 12:34 UTC
i think it's matter of the compiler optimizer: if i have something like f($x+=2, $x+=3); (and comma between arguments is not a sync point, like in C) then the optimizer could collapse $x+=2 and $x+=3 to $x+=5. It's not only matter of order. $_ = 1; f($_+=2, $_+=3); sub f { print shift, ", ", shift, "\n"; } print `perl -v`; __________ 6, 6 This is perl, v5.6.0 built for darwin .... [download] Oha	[reply] [d/l]
Re^5: Puzzling $\| behavior by ikegami (Patriarch) on Oct 08, 2007 at 16:58 UTC
It's not because of an optimization. `# $_ $_[0] $_[1] $_[0] $_[1] # ----- ----- ----- ----- ----- $_ = 1; # 1 do { # local @_; # alias $_[0] = $_+=2; # 3 $_ 3 alias $_[1] = $_+=3; # 6 $_ $_ 6 6 &f; };` [download] Add `$_++;` to `f` and you'll see it print `7, 7` because `$_[0]` and `$_[1]` are aliased to `$_`. It works that way because `+=` returns its LHS as an lvalue. `>perl -le "$_=1; ($_+=2)+=3; print $_" 6` [download]	[reply] [d/l] [select]


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