在同一shell中执行一个块

Question

Im' trying to find a way to execute a block in the same way I would execute a code calling an external script.

Let me exemplify ...

# caller.sh
!#/bin/bash

/soft/executer.sh &

After executing "caller.sh", the "ps" command return will be something like:

PID   TTY      TIME     CMD
19566 pts/7    00:00:00 bash
22689 pts/7    00:00:00 executer.sh
22694 pts/7    00:00:00 ps

But if a change the way to call the script "caller.sh" like this:

# caller.sh
!#/bin/bash

    {
    /soft/executer.sh
    } &

The "ps" command shows up both commands (caller.sh and executer.sh)

PID   TTY      TIME     CMD
19566 pts/7    00:00:00 bash
22689 pts/7    00:00:00 caller.sh
22694 pts/7    00:00:00 ps
22685 pts/7    00:00:00 executer.sh

Both "caller.sh" and "ler.sh" commands are showing up.

I know I could simply use the first option to call this, but this is just a simple example to ask how to unlik the processes "caller.sh" and "execute.sh" in the second example that uses blocks

Thanks

Answer 1

I would try this in caller.sh:

#!/bin/bash

(
    exec /soft/executer.sh
)&

The issue is that a block or subshell is merely a copy of the parent meaning the parent may be gone, but the child has the same name therefore shows up in ps. So, if you have:

    #!/bin/bash

    (
        /soft/executer.sh
    )&
    sleep 60

You will see two copies of caller.sh (the parent and the child). The parent is sleeping and the child is waiting for executer.sh to finish.