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Why does the following print 0? I expected it to print 1234.

$ bash -c exit 1234
$ echo $?
k314159
  • 445

2 Answers2

4
bash -c exit 1234

runs bash with the separate arguments -c, exit, and 1234. -c consumes the argument immediately following it, so bash runs exit only; 1234 is used as the first positional parameter ($0, the name of the shell).

To get bash to exit with a given code, you need to quote the whole command:

bash -c "exit 1234"

Note that bash exit statuses are limited to 0-255, so this produces an exit status of 210 (1234 % 256).

Stephen Kitt
  • 434,908
2

You can't with bash's exit builtin even if you fix your code to run exit 1234 instead of exit alone, as bash takes upon itself to truncate the value to 8 bits:

$ strace -e exit_group bash -c 'exit 1234'
exit_group(210)                         = ?

You need to execute another command in the same process that can exit with arbitrary values:

$ strace -e exit_group bash -c 'exec perl -e "exit 1234"'
exit_group(1234)                        = ?

$ strace -e exit_group bash -c 'exec zsh -c "exit 1234"'
exit_group(1234)                        = ?

In any case, note that some systems like Linux don't let the parent process access the full value of that exit status even when using the waitid() API or a SIGCHLD sigaction() handler which is required by POSIX not to truncate it.

More details on that at:

Stéphane Chazelas
  • 544,893
  • Thanks. I used the number 1234 as just a placeholder, as my puzzle was that it was returning 0 (because of the error in my command line as pointed out by the other answer). I should have just used 123 in my question instead of 1234. But your answer is interesting nevertheless. – k314159 Oct 10 '22 at 09:07