> Yeah. Basically instead of treating ! specially, Bash treats it like any other command that sets $? afterwards. And since it always succeeds, you always end up with $?=0 after it.
I don't think that is true, consider:
# if false ; then echo "true $?" ; else echo "false $?" ; fi
false 1
# if ! false ; then echo "true $?" ; else echo "false $?" ; fi
true 0
# if ! ! false ; then echo "true $?" ; else echo "false $?" ; fi
false 1
Basically, $? evaluates to the full condition inside the "if" sentence.
One of the first things I learned when starting to use $? was to assign it to a variable asap if I needed to use it more than once.
Whoops, that's what I get for writing these when I'm sleepy. Yes of course ! doesn't always set the result to success, that wouldn't make any sense. I misspoke in that part. I meant to say Bash treats it like any other command that modifies $?, instead of treating it specially and preventing it from modifying that variable. Thanks.
I don't think that is true, consider:
Basically, $? evaluates to the full condition inside the "if" sentence.One of the first things I learned when starting to use $? was to assign it to a variable asap if I needed to use it more than once.