Question

当我更改调用程序中的标准文件句柄时，我正在玩shell以及它是如何起作用的。 Proc说：

$ in，$ out和$ err是待启动程序的三个标准流，默认为“ - ”，这意味着它们从父进程继承流。

据我所知，外部程序不使用相同的文件句柄：

#!/Applications/Rakudo/bin/perl6

#`(
    make an external Perl 6 program the outputs to standard handles
    )
my $p6-name = 'in-out.p6'.IO;
#END try $p6-name.unlink; # why does this cause it to fail?
my $p6-fh = open $p6-name, :w;
die "Could not open $p6-name" unless ?$p6-fh;
$p6-fh.put: Q:to/END/;
    #!/Applications/Rakudo/bin/perl6

    $*ERR.say( qq/\t$*PROGRAM: This goes to standard error/ );
    $*OUT.say( qq/\t$*PROGRAM: This goes to standard output/ );
    END
$p6-fh.close;
say $p6-name.e ?? 'File is there' !! 'File is not there';
die "$p6-name does not exist" unless $p6-name.e;

{
#`(
    Start with some messages to show that we can output to
    the standard filehandles.
    )
$*OUT.put: "1. standard output before doing anything weird";
$*ERR.put: "2. standard error before doing anything weird";
shell( "perl6 $p6-name" ).so;
}

{
#`(
    This block assigns a new filehandle to $*OUT and prints a
    message to it. I expect that message to not show up in the
    terminal.

    It then calls run-them to fire off the external process. It
    should inherit the same standard out and its standard out
    messages should not show up. But, they do.
    )
temp $*OUT = open '/dev/null', :w;
$*OUT.put: "3. temp redefine standard output before this message";
shell( "perl6 $p6-name" ).so;
}

$*OUT.put: "4. everything should be back to normal";

输出显示当我打开 / dev / null 并将其文件句柄分配给$*OUT时，当前程序的输出不会显示在终端中（没有输出开始）与3.）。但是，当我调用shell时，其标准输出将转到原始标准输出：

File is there
1. standard output before doing anything weird
2. standard error before doing anything weird
    in-out.p6: This goes to standard error
    in-out.p6: This goes to standard output
    in-out.p6: This goes to standard error
    in-out.p6: This goes to standard output
4. everything should be back to normal

我并不担心如何实现这一目标。我可以创建一个Proc对象并将文件句柄传递给它。

还有其他事情发生吗？

Answer 1

默认情况下，$*OUT中的IO :: Handle绑定到操作系统给出的低级STDOUT文件句柄。

shell和run只是让生成的进程使用为Perl 6提供的低级STDOUT文件，除非另有指定。

Perl 6在产生新流程之前不会改变外部环境。

最简单的方法是使用命名参数将要使用的文件句柄对象提供给shell或run调用。

# no testing for failure because the default is to throw an error anyway

my $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);
}

这导致

STDERR: run
STDOUT: shell

在丢弃输出数据的特定情况下，应使用:!out或:!err。

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

STDOUT: no STDERR

如果您只想为您:out和:err拦截数据，请执行此操作;

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

captured: STDOUT capture

改变Perl 6的$ * OUT是否会改变子进程的标准输出？

1 个答案: