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Re: getting the bash exit code

by Perlbotics (Abbot)
on Aug 24, 2012 at 18:54 UTC ( #989613=note: print w/ replies, xml ) Need Help??


in reply to getting the bash exit code

Now I have no idea how it got 25856 from 101 (well it does seem to be 101 times that magical 256 number, but I'm sure it isn't that simple).

Yes, it's that simple. perlvar explains that $? is set like explained in system where the following snippet is copied from:

Quote from system:

if ($? == -1) { print "failed to execute: $!\n"; } elsif ($? & 127) { printf "child died with signal %d, %s coredump\n", ($? & 127), ($? & 128) ? 'with' : 'without'; } else { printf "child exited with value %d\n", $? >> 8; }
HTH

Update:

Bit# 15 14 13 12 11 10 09 08 07 06 05 04 03 02 01 00 | | | | | | | | | | | | | | | | X7 X6 X5 X4 X3 X2 X1 X0 CD S6 S5 S4 S3 S2 S1 S0 | | | +-- exit code (8bit) | +-- signal code (7bit) | +-- core-dumped flag

Update2 (in response to node below):

So for quick and dirty every day use it sounds like I can just say something like: $exit_status = $?/255;
No.
my $exit_status = int( $? / 256 ); # quick and dirty exit-state # decimal arithmetic so to speak # (works for positive numbers)
Use int() to get rid of potential fractional parts when the process terminated due to a signal or core-dump.
The >>-operator does this in a single step.
I might note, that I assumed a *NIX background. The Windows emulation might not cover the whole interface (exit-code plus signals). Other monks will know better...

So X >> Y is most of the time int( X / (2**Y) ) for positive numbers.

Here are some starters for binary arithmetic: wikibooks, Khan Academy - Binary Numbers, Slideshare - Binary Arithmetic, ... Search also MIT OpenCourseWare - I am sure, they have something in their repertoire.


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Re^2: getting the bash exit code
by xorl (Deacon) on Aug 27, 2012 at 13:22 UTC

    Thanks for the answer. I wish you had stopped at "Yes It's that simple".

    So for quick and dirty every day use it sounds like I can just say something like:
    $exit_status = $?/255;

    I've never understood what bitwise AND (the '&' operator) does. Plus I've never encountered the ">>" operator before (I looked it up it is called a shift operator but the explanation of what it does was way beyond me). From your answer I get the impression I'd need to go back to college and take some pretty advance math or maybe computer science classes to understand them.

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