Does die() change the order of execution?

lo_tech has asked for the wisdom of the Perl Monks concerning the following question:

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;
[download]

Here's the output without an @ARGV:

where are those args again?<br>
-1
[download]

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

Comment on Does die() change the order of execution? Select or Download Code

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Re: Does die() change the order of execution? by runrig (Abbot) on Nov 01, 2001 at 21:50 UTC
STDOUT is buffered and may appear after STDERR. Put this at the top: `$\|++;` [download]	[reply] [d/l]
Re: Does die() change the order of execution? by jwest (Friar) on Nov 01, 2001 at 21:53 UTC
What you're seeing is an illusion. Since you're looking at STDOUT (the target of print) and STDERR (the target of die) at the same time, it looks like that's the order it's executing in. In all actuality, it's happening in the order you expect, but the buffers are being flushed in the "wrong" order, creating the discrepancy. Hope this helps! --jwest -><- -><- -><- -><- -><- All things are Perfect To every last Flaw And bound in accord With Eris's Law - HBT; The Book of Advice, 1:7	[reply]
Re: Does die() change the order of execution? by mkmcconn (Chaplain) on Nov 01, 2001 at 22:00 UTC
runrig gave the answer. Dominus has the explanation. Read his articleSuffering from Buffering. credit goes to blakem for the link to that article mkmcconn	[reply]
Re: Does die() change the order of execution? by Asim (Hermit) on Nov 01, 2001 at 21:58 UTC
Sounds like a buffering issue. Perl buffers the output of commands (i.e. `print` statements using `STDOUT`) automatically, but (as I recall) error messages (using `die` which prints to `STDERR`) come out as soon as possible. Try using: `$\| = 1;` after the `use strict;` and see what comes out. ----Asim, known to some as Woodrow.	[reply] [d/l] [select]