3 ms·

Curious on this one, do you have an example showing it not working as expected?

• dataflow 4y ago
  $ if false; then :; else (if [ $? -eq 0 ]; then echo msg; fi); fi $ if ! false; then (if [ $? -eq 0 ]; then echo msg; fi); else :; fi msg
  • nikau 4y ago
    That's correct behavior though. You should do if ARG=false Then use $ARG not $? in the if statement.
    • dataflow 4y ago
      It's the best kind of correct, yes.
      • nikau 4y ago
        It's a pretty bizarre use case, why would you need the second if statement, you already are in logic based on the return code I can't think of an example of the if statement being confusing that isn't relying on $?
        • dataflow 4y ago
          error=0 if ! some_command; then error=$?; bar; fi echo "blah" return $error
          • nikau 4y ago
            Yep that's a valid scenario, if you need more granularity than just fail or success then you will have to use a different method, e.g. Status=0 Somecommand || status=$? Then do logic based on status
  • 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.
  • Ferret7446 4y ago
    Your original claim is disingenuous. Of course cond = false if cond; then x; else (if cond; then echo msg); fi does not do the same thing as cond = true if cond; then (if cond; then echo msg); else x; fi Bash is not somehow strange in this regard.
    • dataflow 4y ago
      There's neither anything disingenuous about my claim. Bash's behavior is quite unique and unexpected to most people in the examples I showed. And I don't appreciate the attack.