How to execute a subshell directly

Question

I have this:

timeout 25 bash -c '
  for i in {1..9}; do
      if read line < "$my_fifo"; then
          if test "$line" != "0"; then
                   exit 0;
           fi
      fi
  done
'

I don't really like that bash can't do this:

timeout 25 (...)

I don't understand why () is not considered a program by itself. Just an anonymous program... anyway..

my goals is to achieve the above but without the need to use bash -c 'xyz' since I won't get syntax highlighting in the quotes etc.

Is there a workaround for this?

Answer 1

A workaround in Bash may be to define a function, export it, finally use timeout 25 bash -c to run the function. This is even less "directly", but at least syntax highlighting should work.

Frankly, since timeout is a separate program, I think there is no way to execute "directly". timeout 25 (...) would be possible if Bash implemented its own timeout and made it a keyword (like time; time (...) works).

Note in the below example we put $my_fifo in the environment of timeout. Your original code also requires $my_fifo in the environment.

#!/bin/bash
my_func() {
  for i in {1..9}; do
      if read line < "$my_fifo"; then
          if test "$line" != "0"; then
                   exit 0;
           fi
      fi
  done
}
export -f my_func
my_fifo=./fifo timeout 25 bash -c my_func