击子shell /管道 - 这部分在子shell执行？ [英] Bash subshell/pipelines - which parts are executing in subshells?
问题描述
在一<一href=\"http://stackoverflow.com/questions/9108181/append-output-of-one-command-to-the-output-of-another-in-a-single-command/9108410#comment11442552_9108410\">comment在另一个帖子，@JonathanLeffler指出：
In a comment on another post, @JonathanLeffler stated that:
{...} | somecommand在子壳运行，并且不影响
父shell。演示：
{ ... } | somecommand is run in a sub-shell and doesn't affect the parent shell. Demo:
X = PQR;回声$ X; {X = ABC;回声$ X; } |猫;回声$ X
（带输出PQR，ABC，PQR三线）
(with output PQR, ABC, PQR on three lines)
确实
james@bodacious-wired:tmp$X=PQR; echo $X; { X=ABC; echo $X; } | cat; echo $X
PQR
ABC
PQR
然而，男人庆典表示， {...} 不在一个子shell执行：
However, man bash says that { .. } does not execute in a subshell:
{ list; }
list is simply executed in the current shell environment. list must be
terminated with a newline or semicolon. This is known as a group command.
那么是什么回事？为男人庆典错了吗？我知道一个管道的每个部分子shell执行;但我不认为这是如何引起的观察到的行为。例如：
So what's going on here? Is man bash wrong? I know that each part of a pipeline executes in a subshell; but I don't see how that causes the observed behaviour. For instance:
james@bodacious-wired:tmp$X=PQR; echo $X | sed; X=ABC; echo $X | sed; echo $X
PQR
ABC
ABC
编辑补充：
有几个人使用建议回声$$ 来表明事情（或者不是）一个子shell的部分。这是不是在所有有用的，因为 $ 在参数扩展阶段，这恰好长在执行任何命令之前被扩大了。
A few people have suggested using echo $$ to show that things are (or are not) parts of a subshell. This is not at all useful, as the $$ gets expanded during the parameter expansion stage, which happens long before any commands are executed.
作为一个例子：
james@bodacious-wired:tmp$echo 1$$; ps; ( echo 2$$; ps ); echo 3$$; ps
11194
PID TTY TIME CMD
1194 ttys000 0:00.22 -bash
21194
PID TTY TIME CMD
1194 ttys000 0:00.22 -bash
7894 ttys000 0:00.00 -bash
31194
PID TTY TIME CMD
1194 ttys000 0:00.22 -bash
james@bodacious-wired:tmp$
您可以看到 PS 的第二次调用发生子shell里面，具有pid 7894 ;但 2呼应$$ 仍然显示在bash的变量扩张阶段取代的价值，它催生了子shell之前
You can see that the second invocation of ps happened inside a subshell, with pid 7894; but echo 2$$ still shows the value that bash substituted in the variable expansion phase, before it spawned the subshell
有关的对比，并表明 {...} 做的不的产卵子shell：
For contrast, and demonstrating that { .. } does not spawn a subshell:
james@bodacious-wired:tmp$echo 1$$; ps; { echo 2$$; ps; }; echo 3$$; ps
11194
PID TTY TIME CMD
1194 ttys000 0:00.22 -bash
21194
PID TTY TIME CMD
1194 ttys000 0:00.22 -bash
31194
PID TTY TIME CMD
1194 ttys000 0:00.23 -bash
只是为了证明@nos是正确的，管道添加到上面：
Just to prove that @nos is correct, add a pipeline to the above:
james@bodacious-wired:tmp$echo 1$$; ps; { echo 2$$; ps; } | sed ; echo 3$$; ps
11194
PID TTY TIME CMD
1194 ttys000 0:00.25 -bash
21194
PID TTY TIME CMD
1194 ttys000 0:00.25 -bash
7945 ttys000 0:00.00 -bash
7946 ttys000 0:00.00 sed
31194
PID TTY TIME CMD
1194 ttys000 0:00.25 -bash
如所预期的，壳产卵2子shell，一个用于管道的两侧。
As expected, the shell spawns two subshells, one for each side of the pipeline.
推荐答案
一个流水线的每一侧变成一个子shell最少。
Each side of a pipeline becomes a subshell at least.
X = PQR;回声$ X; {X = ABC;回声$ X; } |猫;回声$ X
将会使，ATLEAST {X = ABC子shell /进程;回声$ X; } 和猫。
will make a subshell/process of , atleast { X=ABC; echo $X; } and cat.
管道中的每个命令被作为一个单独的过程（即，在子shell）执行。 ，从人的bash
"Each command in a pipeline is executed as a separate process (i.e., in a subshell)." , from man bash
如果你不是做
X = PQR;回声$ X; {X = ABC;回声$ X; };回音|猫;回声$ X
您会看到这事后回声$ X 显示ABC。
You'll see afterwards that echo $X shows ABC.
有这么命令在子shell执行过其它方式，例如如果背景子命令： {X = SUB;睡眠1; }＆安培; ，该组将在一个子shell中运行，而只是 {X = SUB;睡眠1; } 不会的。
There's other ways that commands are executed in subshells too, e.g. if you background a subcommand: { X=SUB ; sleep 1; } & , that group will run in a subshell, whereas just { X=SUB ; sleep 1; } will not.
如果你想组命令，总是以一个子shell执行，使用括号，（X = ABC;回声$ X）。而不是括号
If you want to group commands that always execute in a subshell, use parenthesis, (X=ABC ; echo $X) instead of braces.
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