Difference between `exit;` and `exit $?;`

Question

Is there any difference between these two commands:

   exec "$(dirname "$0")/suman-shell";
   exit $?;

and

   exec "$(dirname "$0")/suman-shell";
   exit;

is the $? redundant in the first case?

Answer 1

There are several flaws in your script:

exec "$(dirname "$0")/suman-shell";
exit $?;

• The first one is the ending semicolons are redundant. They are harmless but serve no purpose. An end of line is already a separator, semicolons are used as a separator between multiple commands on a single line. Their presence here is slightly worsen the code readability.

• The second one is exit by default returns the exit status of the previous command, so stating $? is redundant, albeit signalling the intent.

• The third one is exec never returns under normal circumstances¹, so the exit call is not reached in the first place and is also redundant.

¹ The only cases where exec returns is an empty argument, a broken redirection or the execfailbash option set (which is not the default setting). The first and second ones do not apply here, and there is no evidence the third one does.

Answer 2

From help exit:

Exits the shell with a status of N. If N is omitted, the exit status is that of the last command executed.

So yes, it is completely redundant.

Answer 3

Strictly speaking, there's a difference between exit and exit $? when $IFS contains digits when you forget to quote that $? and thus are invoking the split+glob operator:

$ bash -c 'IFS=123; false; exit $?'; echo "$?"
bash: line 0: exit: : numeric argument required
2
$ bash -c 'IFS=123; false; exit'; echo "$?"
1
$ bash -c 'IFS=123; false; exit "$?"'; echo "$?"
1
$ bash -c "exit ''"
bash: line 0: exit: : numeric argument required

In the first case, because the $? is unquoted, it's subject to word splitting. With $IFS containing 1, the content of $? (1) is split into one empty argument in bash, so it's as if you had called exit ''.