Re: Passing variables by reference conflicts with local

[Freddy Vulto]

On 100430 08:19, Greg Wooledge wrote:
> On Fri, Apr 30, 2010 at 12:02:58AM +0200, Freddy Vulto wrote:
> > Passing variables by reference however, has a caveat in that
> > local variables override the passing by reference, e.g.:
> > 
> >     t() {
> >         local a
> >         eval $1=b
> >     }
> >     unset a; t a; echo $a  # Outputs nothing, expected "b"
> 
> Why did you declare 'a' to be local if that's not what you wanted?
> Why did you expect 'b' to be output, if you used a local variable?

t() is an example library function which - as a black box - can contain
many local variables.  Local 'a' is just an example.

I would like to call t(), and let it return me a filled variable by
reference, that is without polluting the global environment.

Problem is, all variables work except 'a' because t() happens to declare
this local - which I'm not supposed to know because t() is a black box.

Maybe the example is not clear because 'a' becomes global after all,
and is this a better example:

    # Param: $1  variable name to return value to
    blackbox() {
        local a
        eval $1=bar
    }

This goes all right:

    f() {
        local b
        blackbox b
        echo $b
    }
    f  # Echos "bar" all right

But if I change 'b' to 'a', this conflicts with blackbox() local 'a':

    f() {
        local a
        blackbox a
        echo $a
    }
    f  # Outputs nothing unexpected


Freddy Vulto
http://fvue.nl/wiki/Bash:_passing_variables_by_reference