Re: want to pass vector<foo*> to fn expecting vector<const foo*>

From:
=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 24 Apr 2013 15:05:25 -0700 (PDT)
Message-ID:
<kl95ej$une$1@dont-email.me>
Am 24.04.2013 20:16, schrieb Francis Glassborow:

On 24/04/2013 14:26, Daniel Kr?gler wrote:

On 2013-04-23 06:33, Balog Pal wrote:

The same is true for auto_ptr<foo *> and auto_ptr<const foo *> (and
many other flavors), yet they are implicitly convertible.


Just to get this right: We are talking about conversions here, not
about reinterpretations, right? (These are what the Standard smart
pointers support) If so, I don't understand the concern here, because
it is really not so hard to convert std::vector<foo *> to
std::vector<const foo *> via the iterator-pair constructor, which
supports all implicit conversions, so we can already write

std::vector<foo *> vf = ..;
std::vector<const foo *> vcf(vf.begin(), vf.end());


OK, I am probably losing it, but that creates a copy of vf. Worse than
that the copy has duplicated the pointers and so has the potential for
doing some real damage (remember that destruction is not a mutating
operation, at least it wasn't last time I looked)

The original problem concerned passing by reference which is not a
conversion but an attempt to add an extra constraint (that the pointers
may not be used to mutate the object)


I'm probably losing as well, but my response was inspired by Balog Pal's
comparison:

"The same is true for auto_ptr<foo *> and auto_ptr<const foo *> (and
many other flavors), yet they are implicitly convertible."

This model adapted to std::vector<foo *> and std::vector<const foo *>
looks comparable to me.

Greetings from Bremen,

Daniel Kr?gler

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