Code block usage { } in bash

Question

I was wondering why code block was used in this example below:

possibly_hanging_job & { sleep ${TIMEOUT}; eval 'kill -9 $!' &> /dev/null; }

This could have been written like this ( without using code block as well) ..right ?

possibly_hanging_job & 
sleep ${TIMEOUT}
eval 'kill -9 $!' &> /dev/null

Answer 1

Putting the last two commands in braces makes it clear that “These are not just two additional commands that we happen to be running after the long-running process that might hang; they are, instead, integral to getting it shut down correctly before we proceed with the rest of the shell script.” If the author had instead written:

command &
a
b
c

it would not be completely clear that a and b are just part of getting command to end correctly. By writing it like this:

command & { a; b; }
c

the author makes it clearer that a and b exist for the sake of getting command completely ended and cleaned up before the actual next step, c, occurs.

Answer 2

Actually I even wonder why there's an eval. As far as I see it should also work without that.

Regarding your actual question: I guess the code block is there to emphasize that the sleep belongs to kill. But it's not necessary. It should also work like this:

possibly_hanging_job & sleep ${TIMEOUT}; kill -9 $! &> /dev/null