Does changing Perl 6's $*OUT change standard output for child processes?

By default the IO::Handle that is in $*OUT is bound to the low-level STDOUT filehandle given by the operating system.

shell and run just let the spawned process use the low-level STDOUT file that was given to Perl 6, unless you specify otherwise.

Perl 6 doesn't change anything about the outside environment until the moment before it spawns a new process.

The simplest thing to do is to give the filehandle object you want to use to the shell or run call with a named argument.

# no testing for failure because the default is to throw an error anywaymy $p6-name = 'in-out.p6'.IO;END $p6-name.unlink;$p6-name.spurt(Q'put "STDOUT: @*ARGS[0]";note "STDERR: @*ARGS[0]"');run $*EXECUTABLE, $p6-name, 'run', :out(open '/dev/null', :w);{  temp $*OUT = open '/dev/null', :w;  shell "'$*EXECUTABLE' '$p6-name' 'shell'", :err($*OUT);}

This results in

STDERR: runSTDOUT: shell

In the particular case of throwing away the output data, :!out or :!err should be used instead.

run $*EXECUTABLE, $p6-name, 'no STDERR', :!err;

STDOUT: no STDERR

If you just want the data to be intercepted for you :out and :err do just that;

my $fh = run( $*EXECUTABLE, $p6-name, 'capture', :out ).out;print 'captured: ',$fh.slurp-rest;

captured: STDOUT capture