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Re: [BUG] 'unset' fails silently under specific conditions

From: Martijn Dekker
Subject: Re: [BUG] 'unset' fails silently under specific conditions
Date: Wed, 2 May 2018 15:07:42 +0100
User-agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.11; rv:52.0) Gecko/20100101 Thunderbird/52.7.0

Op 02-05-18 om 02:20 schreef Chet Ramey:
You complained that `typeset +x' didn't `unexport' a variable.  The reason > is 
that the variable assignment preceding the special builtin caused a
variable to be created at the global scope, and the `typeset' resulted in
a local variable being created.

I still can't see how that is relevant here: 'typeset'/'declare' is not involved in the current issue. But then, neither is 'unset'. It does appear that the current issue is not what I thought it was.

Instead, in POSIX mode, it looks like a variable assignment preceding a special builtin may create a variable at a function-local scope without 'typeset'/'declare' being involved at all. But not always.

Let's see if I'm getting it right this time. In the following:
        set -o posix
        f() { foo=bar : ; }
the command 'foo=bar :'
1. makes the assignment 'foo=bar' survive the ':' command;
2. gives 'foo' a global scope;
3. permanently exports the global variable 'foo'.

However, in the following:
        set -o posix
        f() { foo=bar; foo=baz : ; }
the plain assignment 'foo=bar' creates an ordinary global variable named 'foo' with value 'bar', and then the command 'foo=bar :' 1. makes the assignment 'foo=baz' survive the ':' command, but by creating *another* 'foo' instead of overwriting the first;
2. gives that other 'foo' a function-local scope;
3. exports the local variable 'foo' for the duration of its existence.

My testing confirms that this is what appears to happen, and I think it's a bug, because (1) POSIX has no notion of variables with a function-local scope; (2) even if it did, no command was issued that should make the variable local to the function; and (3) the behaviour in the second example is inconsistent with that in the first.

I think 'foo=baz :' should act the same in the second example as 'foo=bar :' does in the first, i.e.: if there is already a variable by that name it should be overwritten, just like with any normal shell assignment.

- M.

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