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From: John E. Potter (jpotter_at_[hidden])
Date: 2001-06-07 10:14:35

On Thu, 7 Jun 2001, David Abrahams wrote:

> From: "John E. Potter" <jpotter_at_[hidden]>

> > Not a problem for a language with modifiable rvalues. Operator= is
> > a member.
> I often forget whether it's classes or built-in types that can be modified
> in that way. When it's a member function, you're not binding X(a) to a
> non-const reference?

Right. I can neither bind an rvalue to a non-const reference nor take
its address, but the implementation can magically produce an rvalue
called this which acts like a poiinter to the non-const rvalue.

> Is this the standard's way of saying "use a self-proxy"?

I think that it is the standard's way of saying that all known output
iterators at the time were self-proxies. I can write an iterator that
will work in all algorithms which specify an output_iterator yet not
meet the requirements for an output iterator. Note also that no
standard iterator of any other type meets the requirements of an
output iterator. For a pointer, I must say *p = t, but the standard
requires p = t. Is it a defect or does it pass because p = t is
equivalent to P(p) = t, they are both syntax errors? Is the intent
to require a = t or just allow it? I think the intent was to say
that all copies of an output iterator are equivalent, but I don't
see any reason to require that either, but it should be allowed.

> The question remains, however: what is the type of t?

Anything which may be used in the expression *a = t. There could be
many different types. You don't want me writing
my_cute_ostream_iterator with operator= for char, int, and double?
Note that for ostream_iterator<int> it, *it = 5, *it = 'X', *it = 3.14
are equally valid. My_cute_ostream_iterator would produce nicer

I once wanted the value_type of an output iterator in an algorithm,
but knowing what the algorithm did and the value_type of the
input iterator allowed me to declare a local of an appropriate
type. It might not be the same as the output iterator, but the
assignment would work.


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