bashprompt$ ./program.pl | ./program.pl — nothing could be simpler...

Course, I went and assumed you meant bash. If you meant open, then you want: open my $out, "|", "./program.pl" or die "lol: $!".

-Paul

In reply to Re: pass variable to another process when invoking from command line by jettero
in thread pass variable to another process when invoking from command line by vineet2004

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