I'm getting to a point of frustration with a rather simple concept, so I seek the wisdom of those that spend more time with perl than I do. The basic idea is that my perl script's STDOUT is not a tty (reasons unimportant), and my perl script is going to call an external program via system(), which checks to see if STDOUT is a tty (using isatty()) and changes its behavior accordingly. I want to fake the external program out and convince it that STDOUT is a tty, when it is in fact not, but I still want the output to end up going to my STDOUT.
Here's code from my last attempt to get this to work, which I think will probably illuminate whatever it is I'm failing to understand about pseudo-ttys and perl filehandles and whatnot. It attempts to create a pseudo-tty, and then use fdopen to "inject" the pseudo-tty into the path of STDOUT traffic to make STDOUT look like a tty:
use IO::Pty;
use IO::Handle;
open(REAL_STDOUT,">&STDOUT");
my $pty = new IO::Pty;
my $slave = $pty->slave;
STDOUT->fdopen($slave,'w') || die $!;
$pty->fdopen(\*REAL_STDOUT,'w') || die $!;
system("echo This is stdout output from an external program");
/Edited to add more...
Round 2: Some more investigation and now I'm at this solution (stolen from some webpage somewhere and hacked up). This seems to get the output through the pty, but the while/print loop at the bottom never terminates, even though the process writing to the filehandle in question is long gone...
use IO::Handle;
use IO::Pty;
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+
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sub do_cmd() {
my $pty = new IO::Pty;
defined (my $child = fork) or die "Can't fork: $!";
return $pty if $child;
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POSIX::setsid();
my $slave = $pty->slave;
close($pty);
STDOUT->fdopen($slave,'>') || die $!;
STDERR->fdopen(\*STDOUT,'>') || die $!;
system("echo This is stdout output from an external program");
exit 0;
}
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my $fh = do_cmd();
while(<$fh>) { print; }