Re: Bug in function return statement in while subshell

[Roman Rakus]

On 07/30/2013 03:50 PM, Chet Ramey wrote:
> On 7/29/13 1:15 PM, Roman Rakus wrote:
> > On 07/29/2013 05:06 PM, Chet Ramey wrote:
> > > On 7/29/13 10:55 AM, Roman Rakus wrote:
> > > > I didn't take a look on where the problem could be, but it is discussed on
> > > > stackoverflow [1].
> > > >
> > > > Looks like return builtin falsely exit execution of while loop instead of
> > > > function.
> > >
> > > What would you like to see happen?  You're in a subshell: the function
> > > can't return, since this is not the shell that called it.  Do you want
> > > parent and child shells both continuing execution after the function call?
> > >
> > > Chet
> > As Chris said, some error, because return is not in function.
> >
> > Example:
> > f1() {
> >      : | while :; do return 3; done
> >      echo $?
> >      return 1
> > }
> > echo $?
> > f1; echo $?
> > return 2
> >
> > Looks like the bash doesn't restore the i'm-in-function indicator when
> > running compound command (like while or if) in functions' subshell.
>
> I'm not sure exactly what you mean by this, since the `return' call in
> the pipeline inside the function completes without an error.  Bash
> prints an error message due to the `return 2' not occurring while in
> a function.
>
> Chet
My point is that the `return 3' in while loop is in subshell and will 
not return from the function and it can be perceived similarly to 
`return 2' statement. Why the `return 3' doesn't introduce an error? It 
is not returning from a function. It just naively knows that it is in 
function?

RR