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; + + + sub do_cmd() { my $pty = new IO::Pty; defined (my $child = fork) or die "Can't fork: $!"; return $pty if $child; + 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; } + my $fh = do_cmd(); while(<$fh>) { print; }

In reply to How do you properly tty-ify a non-tty STDOUT? by ph713

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