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From: Keith Thompson <kst-u@mib.org>
Newsgroups: comp.lang.c
Subject: Re: C++ features for C
Date: Sun, 13 May 2018 14:10:23 -0700
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David Brown <david.brown@hesbynett.no> writes:
[...]
> Going back to your "double x;" example, "&x" does /not/ "yield a double* 
> pointer".

I'm not sure what you're trying to say, but yes, it does.  More
precisely, evaluating the expression `&x` yields a value of type
`double*`.

>            "&x" is an /expression/ of type "double * const".  There is 
> no pointer.

There is no const.  &x is of type double*, which is a pointer type.  (&x
is not an lvalue, so of course its value can't be modified, but "const"
isn't used that way).

I think the point you're making is that given
    double x;
there is no pointer *object*.  Evaluating `&x` does yield a pointer
*value*.  (Using pointer as a noun rather than as an adjective can lead
to a lot of confusion.)

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
Working, but not speaking, for JetHead Development, Inc.
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"