Re: [BUG] persistently assigned variable cannot be unexported in POSIX mode

[Martijn Dekker]

Op 27-04-18 om 22:16 schreef Chet Ramey:
> On 4/25/18 10:51 PM, Martijn Dekker wrote:
>
> > What I'm reporting here is a bug I discovered with unexporting a variable
> > that is so exported while bash is in POSIX mode. It cannot be unexported
> > using 'typeset +x' if you try to do that in a shell function.
> >
> > This works:
> >
> > $ bash -o posix -c 'foo=abc : ; typeset +x foo; env|grep ^foo='
> > (no output, as expected: no longer exported)
> >
> > But this doesn't:
> >
> > $ bash -o posix -c 'fn() { foo=abc : ; typeset +x foo; env|grep ^foo=; }; fn'
> > foo=abc
>
> It seems like you're assuming that in posix mode, variable assignments that
> precede special builtins executed in shell functions should create local
> variables. Is that correct?

No. The issue is:

1. In POSIX mode, as ':' is a so-called special builtin, "foo=abc :" 
causes the assignment to 'foo' to persist past the ':' command, and 
'foo' also remains exported. This is all POSIX-compliant.

(I really wish it didn't remain exported, though. POSIX allows it, but 
doesn't mandate it; it's unspecified whether it remains exported or not. 
But it seems broken for it to remain exported. In fact I'm not sure why 
it's exported to begin with. As far as I know, there are no special 
builtins that run external binaries.)

2. The bug is: 'declare +x' a.k.a. 'typeset +x' then fails to unexport 
the variable in the second version above. The variable remains exported 
past 'typeset +x foo', as proven by grepping the output of 'env'.

It shouldn't matter if the variable is local to the function or not, as 
everything is done in the same scope. 'declare +x' should clear the 
export flag as it is documented to do. In the second invocation quoted 
above, it doesn't.

Having said that, I now cannot reproduce the issue in the 2018-04-27 
development snapshot... so thanks for fixing it anyhow :)

- Martijn