30 Mayıs 2020 Cumartesi tarihinde Laurent Picquet <lpicq...@gmail.com> yazdı:
> Hello Dale, > > This is really interesting. > Should the 'local' command be the one able to detect that the assignment to > the variable had an non-zero exit code and return the non-zero exit code? > > as a developer, it is counter-intuitive that the 'local' command tells us > everything is ok when it wasn't. If feel it should know that the assignment > encountered a problem and should report it > > Everything is ok for `local` though; it takes a valid assignment statement and successfully evaluates that. So it's not that the assignment encountered a problem, but that the expansion has failed, which has nothing to do with `local`. So there is no reason for `local` to return a non-zero exit status in that case. > The return status is zero unless local is used outside a function, an > invalid name is supplied, or name is a readonly variable. > > > > On Fri, 29 May 2020 at 03:43, Dale R. Worley <wor...@alum.mit.edu> wrote: > > > It's a subtle point. See this paragraph in the bash manual page: > > > > If there is a command name left after expansion, execution > > proceeds as described below. Otherwise, the command exits. If > > one of the expansions contained a command substitution, the exit > > status of the command is the exit status of the last command > > substitution performed. If there were no command substitutions, > > the command exits with a status of zero. > > > > In one of your examples, a "local" command is generated using a command > > substitution, so the exit status is that of the local command. In the > > other, only an assignment is done, which is not a command, so the exit > > status is that of the last command substitution. > > > > Dale > > > > > -- > > > -- > > Laurent Picquet > > 16, Hunters Chase > > South Godstone > > RH98HR > > England > > tel: 07882 356 104 > -- Oğuz