How to properly capture the return value of a unix command?

Question

I'm having trouble getting the return value of a unix command into a perl variable.

Unix command:

#nc -z 8.8.8.8 441; echo $?
1

Perl command:

#perl -e 'my $pstate=`nc -z 8.8.8.8 441; echo $?`; print $pstate;'
0

So the perl command seems to get a return value of "no error"? How can I properly capture the return value of the *nix command?

Another instance:

#perl -e 'my $pstate=`ping -v 8.8.8.8 -c 1`; print $pstate;'
PING 8.8.8.8 (8.8.8.8) 56(84) bytes of data.

This returns the proper value. So what am I doing wrong in the first instance?

Answer 1

Variables are interpolated inside backticks, so the $? in

my $pstate=`nc -z 8.8.8.8 441; echo $?`

refers to Perl's $?, not the shell's $?. And what the shell sees is something like

nc -z 8.8.8.8 441 ; echo 0

To fix this, you can escape the shell command

my $pstate=`nc -z 8.8.8.8 441; echo \$?`;

or use the qx operator with the single quote as a separator (this is the one exception to the "interpolation inside the qx operator" rule)

my $pstate=qx'nc -z 8.8.8.8 441; echo $?';

or use readpipe with a non-interpolated quote construction

my $pstate= readpipe( 'nc -z 8.8.8.8 441; echo $?' );
my $pstate= readpipe( q{nc -z 8.8.8.8 441; echo $?} );