On 4/27/05, John W. Krahn <[EMAIL PROTECTED]> wrote:
Jay Savage wrote:
4> open (V4, "samcmd v4 2>\&1 |" ) || die "unable to open pipe... Broken?$!";
Don't do this. the precedence of || is too high. your code attempts to open a pipe, and if it can't, then it attempts to open "die..." and starts throwing exceptions.
No, that is NOT what happens, it will NEVER attempt to open "die..." with or without the parentheses. The ONLY time it will attempt to open "die..." is if there are no parentheses and the expression on the left hand side of the || operator evaluates to false.
open V4, '0' || die $!;
But even then it will NOT attempt to open "die..." because die() exits the program!
Is stand corrected. There is no exception; I guess any time I've run
into it, I've relied on whatever was opened, and died anyway. But I
don't know what else to call opens behavior, except attempting to open
die. Except in the case of parenthesis, as you noted, the behavior of
open || die sure looks like this to me: open (X, badfile || die). The only difference between the two expressions below is the
precedence of the operator.
[EMAIL PROTECTED]:~> perl -e 'open FH, "< BAdFiLe" || die "$!"' [EMAIL PROTECTED]:~> perl -e 'open FH, "< BAdFiLe" or die "$!"' No such file or directory at -e line 1.
Open may not technically be trying to open an expression and failing, I don't know. To be honest, I've never taken apart the source to see. But the appearance is certainly that that's what happens, and the result is so similar as to not matter. Especially consider the following:
perl -e 'open FH, "< BAdFiLe" || die or die "$!"; print "didnt die\n" ' No such file or directory at -e line 1.
Where did the first die go if || didn't attempt to pass it to open?
if the reason for the failed open were the attempt to open "BAdFiLe",
the first die would execute and the program would exit bore it got to
the second. But clearly that's not what's happening. The first die
is getting slurped up by ||, which is presumably trying to pass it on
to open. When that fails, the second die executes, exiting with $!. at least that's what it looks like to me.
So what's really happeneing here?
perldoc perlop
[snip]
C-style Logical Or
Binary "||" performs a short-circuit logical OR operation. That is, if
^^^^^^^^^^^^^
the left operand is true, the right operand is not even evaluated.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^Perhaps this will help illustrate:
$ perl -MO=Deparse,-p -e'open IN, "BADFILE" || die "ERROR: $!"'
open(IN, 'BADFILE');
-e syntax OK
$ perl -MO=Deparse,-p -e'open IN, "BADFILE" || die or die "ERROR: $!"'
(open(IN, 'BADFILE') or die("ERROR: $!"));
-e syntax OKBecause the string "BADFILE" is true, which is determined at compile time, it is as if "|| die" did not exist at all!
Of course if the file name is in a variable then it must be evaluated at run time:
$ perl -MO=Deparse,-p -e'open IN, $ARGV[0] || die "ERROR: $!"' BADFILE
open(IN, ($ARGV[0] || die("ERROR: $!")));
-e syntax OK
$ perl -MO=Deparse,-p -e'open IN, $ARGV[0] || die or die "ERROR: $!"' BADFILE
(open(IN, ($ARGV[0] || die)) or die("ERROR: $!"));
-e syntax OKWhere "|| die" will only be evaluated if the variable is false (undef, 0, '0' or '').
John -- use Perl; program fulfillment
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