Why is the exit status always 0 at the start of a script?

Question

I have a script like this, named judge:

#!/bin/bash
echo "last exit status is $?"

It always outputs "last exit status is 0". Eg:

ls -l;   judge # correctly reports 0
ls -z;   judge # incorrectly reports 0
beedogs; judge # incorrectly reports 0

Why?

Answer 1

There are different bash processes executing each line of code and $? isn't shared between the processes. You can work around this by making judge a bash function:

[root@xxx httpd]# type judge
judge is a function
judge ()
{
    echo "last exit status is $?"
}
[root@xxx httpd]# ls -l / >/dev/null 2>&1; judge
last exit status is 0
[root@xxx httpd]# ls -l /doesntExist >/dev/null 2>&1; judge
last exit status is 2
[root@xxx httpd]#

Answer 2

As discussed in the comments, the $? variable holds the value from the last process that returned a value to the shell.

If judge needs to do something based on a previous command state, you could have it accept a parameter, and pass in the state.

#!/bin/bash
echo "last exit status is $1"
# Or even 
return $1

So:

[cmd args...]; judge $?