How to get the perl exit value from bash script

Question

I'm trying to run perl script through bash, and get the perl's exit value.

perl_script.pl

print "test1";
sub a{
  my @array = ("a","b");
  if ($#array ne -1){
   return 1;
  }
  else {return 0;}
}
my $result=a(arg1,arg2);
exit $result;

bash.sh

VARIABLE_1=$("perl_script.pl" arg1 arg2)
RESULT=$?

The '$?' variable keeps returning 0, no matter the exit value is. Do you know another way to retrieve the perl exit value from bash?

Thanks in advance!

A few thoughts on that: add a shebang to your perl script, make it executable, use `./perl_script.pl` — Cyrus, May 01 '20 at 19:06
Your bash code is correct. Something else is going wrong or missing from your question — jordanm, May 01 '20 at 19:13
@Shieryn I copied your perl and bash exactly as it is and did not reproduce the issue. — jordanm, May 01 '20 at 19:40
Tip: `ne` peforms a string comparions. You should use `!=` to compare numbers. — ikegami, May 01 '20 at 19:44
Tip: `if ($#array != -1)` is equivalent to `if (@array != 0)` which is equivalent to `if (@array)` — ikegami, May 01 '20 at 19:45
Can't get this to behave like you describe, in any variation I can think of. Something in your real code is different enough --- please check _carefully_. One thing: add `use warnings; use strict;` to the top of your Perl script so that it alerts you to errors/typos/... (should _always_ have them). — zdim, May 01 '20 at 20:49

ikegami · Answer 1 · 2020-05-01T20:08:18.060

1

bash's $? will be set to the value passed to exit^[1].

$ perl -e'exit 3'

$ echo $?
3

$ perl -e'exit 4'

$ echo $?
4

$ perl perl_script.pl
test1
$ echo $?
1

If the program dies from an exception, the exit code will be set to a non-zero value. If the program neither dies nor calls exit, the exit code will by set to zero.

edited May 01 '20 at 20:08

answered May 01 '20 at 19:44

ikegami

367,544
15
269
518

score -1 · Answer 2 · answered May 01 '20 at 20:24

-1

I noticed the root cause of this problem. It is a simple issue but tricky. So I would like to share it here.

bash.sh

VARIABLE_1=$("perl_script.pl" arg1 arg2)
echo $?
RESULT=$?

The echo process has accessed the $?, so the RESULT variables would save the exit value of the echo operation which always turns 0 (successful operation)

The fix

VARIABLE_1=$("perl_script.pl" arg1 arg2)
RESULT=$? 
echo $RESULT

As the conclusion, the $? has one-time use only right after the script executed.

answered May 01 '20 at 20:24

Shieryn

234
2
15

Nit: `echo` can exit with a code other than 0. [Example](https://pastebin.com/cdc92cJa) – ikegami May 01 '20 at 20:28
Why did you leave the echo out from the question? – jordanm May 01 '20 at 20:28
5

In other words, please don't say the code you posted does something when it's doesn't, and please only post code that exhibits the behaviour you describe. – ikegami May 01 '20 at 20:30
So ... do you actually have an extra `echo $?` in your real code (not shown in Q)? If this is the case I suggest to update the question with a clear added **Note** about it. Or, are you just posting an observation? Either way, this isn't an "answer" and I suggest you remove it. As for the [`$?` variable](https://www.gnu.org/software/bash/manual/bash.html#Special-Parameters), it's always set by the shell for the last executed command, loosely speaking. See for instance [this post](https://stackoverflow.com/questions/6834487/what-is-the-dollar-question-mark-variable-in-shell-scripting). – zdim May 02 '20 at 01:05

How to get the perl exit value from bash script

2 Answers2