Perl: open/close capturing return code - perl

Question

myb.py
import time
import sys
stime = time.time()
run_until = 600
cnt = 0
while True:
dur = time.time() - stime
if dur > run_until:
break
cnt += 1
print cnt
time.sleep(1)
if cnt == 10:
sys.exit(2) <---- capture 2
mya.pl
use FileHandle;
my $myexe = 'myb.py';
my $FH = FileHandle->new;
open $FH, q{-|},
"$myexe 2>&1"
or print "Cannot open\n";
process_output($FH);
close $FH or warn $!;
sub process_output {
my ($fh) = #_;
while (my $line = <$fh>) {
chomp $line;
print "$line\n";
}
}
OUTPUT:
1
2
3
4
5
6
7
8
9
10
Warning: something's wrong at ./mya.pl line 10.
if i change the line to:
my $err = close $FH;
it gives me a blank for $err.
Question: How can I capture the return code 2 from myb.py in mya.pl?

Answer 1

As documented in http://perldoc.perl.org/functions/close.html, the exit value is available as part of $?. But it can be more convenient to use a wrapper:
use IPC::System::Simple qw(capture $EXITVAL EXIT_ANY);
my #output = capture([0,2], "$myexe 2>&1");
print #output;
print "Program exited with value $EXITVAL\n";
The [0,2] says that exit values 0 or 2 are expected, and anything else is a fatal error; you can use EXIT_ANY instead.
This does get all the output at the end, rather than when it is produced, though.

Answer 2

When open creates a child, close functions as waitpid and sets $? accordingly.
$ perl -e'
open(my $fh, "-|", #ARGV)
or die $!;
print while <$fh>;
close($fh);
if ($? == -1 ) { die $!; }
elsif ($? & 0x7F) { die "Killed by signal ".($? & 0x7F)."\n"; }
elsif ($? >> 8 ) { die "Exited with error ".($? >> 8)."\n"; }
' perl -E'
$| = 1;
for (1..5) {
say;
sleep 1;
}
exit 2;
'
1
2
3
4
5
Exited with error 2