Re: About the "nameref" feature in bash...

In article <vj48k5$3r1bq$1@dont-email.me>,
Janis Papanagnou  <janis_papanagnou+ng@hotmail.com> wrote:
...

This is all well and good, but it leaves open the question of "How do I
assign a value to the nameref itself, given that any reference to the
nameref gets converted to a reference to whatever the nameref points to?"

>
If used in functions - where the actual function parameter from the
call cannot be changed from within the function ex post - it makes
sense to not "overwrite" the calling parameter.

To clarify, my actual use case does not involve functions at all.  That's
why, in my OP, I made it clear that "the docs" talk mostly in terms of
using it with functions - implying that that was not my use case.

My use case is in mainline code, where I need to load up a bunch of
shell variables from a database.  Like this:

    mapfile -t < <(sqlite3 ...)
    idx=0
    for MyNameRef in var1 var2 var3 ...
    do MyNameRef="${MAPFILE[idx++]}"
    done

It all works fine; it all works just like the docs say it would.

I just find it a little strange; that's all.  There is another (non-Unix)
scripting language with which I am familiar, where they implemented
"pointers" late-in-development; they followed the C idiom more closely,
where an unadorned instance of the pointer variable refers to the pointer
itself and to dereference it, you stick a * in front.  This may be a little
more verbose, but it makes more sense to me.

Anyway, it seems clear at this point that the only 2 ways to assign to the
pointer itself (in "mainline code") are as described in the OP.  I'm OK
with that.

--
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