Capture exit code of exit command

If you have a script that runs some program and looks at the program's exit status (with $?), and you want to test that script by doing something that causes $? to be set to some known value (e.g., 3), just do

(exit 3)

The parentheses create a sub-shell.  Then the exit command causes that sub-shell to exit with the specified exit status.

exit is a bash built-in, so you can't exec it. Per bash's manual:

Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.

Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS when appropriate.

You can write a function that returns the status given as argument, or 255 if none given. (I call it ret as it "returns" its value.)

ret() { return "${1:-255}"; }

and use ret in place of your call to exit. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.

Some measurements.

time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'

on my machine takes about 3.5 seconds.

 time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'

on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be : or true then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;: to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret solution is over 100 times faster.

Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems. 0.0