Does die() change the order of execution?

[lo_tech]

Could anyone explain this to me?

Here's the code:

#!/usr/bin/perl -Tw
use strict;

print $#ARGV;
die "where are those args again?\n" unless ($#ARGV > 0);
exit;

Here's the output without an @ARGV:

where are those args again?<br>
-1

Here's the question:
Why is the print after the die?

[runrig]

STDOUT is buffered and may appear after STDERR. Put this at the top:

$|++;

[jwest]

What you're seeing is an illusion. Since you're looking at STDOUT (the target of print) and STDERR (the target of die) at the same time, it looks like that's the order it's executing in. In all actuality, it's happening in the order you expect, but the buffers are being flushed in the "wrong" order, creating the discrepancy.

Hope this helps!

--jwest



-><- -><- -><- -><- -><-
All things are Perfect
    To every last Flaw
    And bound in accord
         With Eris's Law
 - HBT; The Book of Advice, 1:7

[mkmcconn]

runrig gave the answer.
Dominus has the explanation. Read his articleSuffering from Buffering.

credit goes to blakem for the link to that article
mkmcconn

[Asim]

Sounds like a buffering issue. Perl buffers the output of commands (i.e. print statements using STDOUT) automatically, but (as I recall) error messages (using die which prints to STDERR) come out as soon as possible. Try using:
$| = 1;
after the use strict; and see what comes out.

----Asim, known to some as Woodrow.