3 ms·

Yep as a programing language shell is weird, I am sure you know this better than me, but I got nerd sniped by your snippet, and want say what I think is happeni

by somat 4y ago

Yep as a programing language shell is weird, I am sure you know this better than me, but I got nerd sniped by your snippet, and want say what I think is happening. false the command returns 1, ! the command turns a 0 return code into a 1 and anything other than a zero into a 0, if is vaguely backwards from most languages in that it deals in return codes a rc of 0 is true and anything else is false in the first case false has a rc of 1 this will execute the else list then the test command 1 is equal to 0(rc = 1 false list) so "msg" does not print in the second case false(rc 1), !(rc 0), if(true list) now you check if rc is 0, it is and "msg" is printed.

dataflow 4y ago
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 know what kind of shell scripts the designers of this write, but the number of times I've wanted or found it helpful for ! to modify $? is has been exactly zero. I am almost certain that >99% of people who write shell scripts do not find this intuitive.
- trashtester 4y ago
  > 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.
  - dataflow 4y ago
    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.

by somat 4y ago

dataflow 4y ago

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 know what kind of shell scripts the designers of this write, but the number of times I've wanted or found it helpful for ! to modify $? is has been exactly zero. I am almost certain that >99% of people who write shell scripts do not find this intuitive.

trashtester 4y ago
> 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.
- dataflow 4y ago
  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.