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Re: [BUG] Can't mark unset variables as read-only

On 4/29/15 10:41 AM, Peter Stephenson wrote:
> On Wed, 29 Apr 2015 06:55:56 -0700
> Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx> wrote:
>> On Apr 29, 11:36am, Peter Stephenson wrote:
>> }
>> } +Note that in zsh (unlike other shells) it is still possible
>> } +to create a local variable of the same name as this is considered a
>> } +different variable (though this variable, too, can be marked readonly).
>> Hrm, I wonder if that, combined with "typeset" automatically creating
>> locals when called in a function context (also a POSIX incompatibility?) 

It's not a Posix incompatibility; Posix does not specify typeset or local
variables at all.  The Posix-conformant ways of declaring variables in a
function result in global variables, as Posix specifies.

>> defeats the whole purpose of an unset readonly?

The question is whether or not you can use local variables to `shadow'
readonly variables in a previous scope.  Bash doesn't allow you to do
that.  Given this script:

readonly foo=immutable

	local foo=bar
	echo inside: $foo

echo outside: $foo

Bash gives the following output:

./x26: line 5: local: foo: readonly variable
inside: immutable
outside: immutable

If the purpose of readonly is to make a particular name/value pair
immutable, I think that allowing a local variable to shadow it,
especially if you're going to export that local variable, is a bad thing.

> I don't think so... bash makes these local, too:
> $ foo=bar
> $ fn() { typeset foo=something_else; echo $foo; }
> $ fn
> something_else
> $ echo $foo
> bar
> So "foo" in fn is logically something entirely different and I don't
> think there's any requirement for it to be treated as readonly;

There's no requirement, since there are no standards for local variables
or variable scoping.  Bash does what I think is right.

``The lyf so short, the craft so long to lerne.'' - Chaucer
		 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU    chet@xxxxxxxx    http://cnswww.cns.cwru.edu/~chet/

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