Re: When does bash spawn a sub-shell with ( list ) ?

[Chet Ramey]

On 1/27/22 2:45 PM, Masahiro Yamada wrote:
> Hi,
>
> The bash manual [1] says
> "Placing a list of commands between parentheses causes
> a subshell environment to be created"
>
> In this statement, "a list of commands"
> means "two commands or more",
> excluding a single command.
>
> Is this correct?

No. Bash will sometimes optimize away a child process creation (in effect,
turning `( foo )' into `(exec foo)') if it can determine that the command
has no side effects. You're seeing that here.

> masahiro@grover:/tmp/test$ cat test.c
> #include <stdio.h>
> #include <unistd.h>
>
> int main(void)
> {
>      printf("pid=%d, ppid=%d\n", getpid(), getppid());
> }
>
> masahiro@grover:/tmp/test$ gcc test.c
> masahiro@grover:/tmp/test$
> masahiro@grover:/tmp/test$ echo $$
> 1071740
> masahiro@grover:/tmp/test$ ./a.out
> pid=1073015, ppid=1071740
> masahiro@grover:/tmp/test$ ( ./a.out )
> pid=1073016, ppid=1071740

The shell optimized away the second fork; the command is still being
executed in a subshell environment and actions by ./a.out will not
affect its parent.

> masahiro@grover:/tmp/test$ ( ./a.out; ./a.out )
> pid=1073020, ppid=1073019
> pid=1073021, ppid=1073019

The shell couldn't optimize it away (it probably could with the last
command in the list, but it doesn't try).

> The command
>      ./a.out
> and
>      ( ./a.out )
> printed the same ppid, 1071740.
> So,  ( ./a.out ) did not fork a new subshell.

It did. Then it ran ./a.out in that subshell without forking again.
Why do you need the shell to fork twice?

--
``The lyf so short, the craft so long to lerne.'' - Chaucer
                 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU    chet@case.edu    http://tiswww.cwru.edu/~chet/