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From: Emil Dotchevski (emildotchevski_at_[hidden])
Date: 2019-12-03 04:36:51

Right, that's what I meant. Dangers, issues with auto aside, I can't think
of a use case when I would want op+ to return a string of size M+N. Keep in
mind, op+ composes. Leaving it undefined is an option, I suppose.

On Mon, Dec 2, 2019 at 7:59 PM Gavin Lambert via Boost <
boost_at_[hidden]> wrote:

> On 3/12/2019 15:46, Emil Dotchevski wrote:
> > I wonder if it is possible to implement op+ such that if one writes:
> >
> > x = a+b;
> >
> > the return value of a+b gets converted to the type of x. The motivation
> is
> > that, as a user, I've determined a fixed size limit for x, and therefore
> > the library should use that, and if it is exceeded at run-time, it makes
> > sense for the behavior to be the same as when any other fixed_string
> > capacity is exceeded (e.g. boost::throw_exception).
> It is possible, by using a deferred execution pattern.
> operator+ would return a unique type which contains the references to a
> and b (not yet concatenated) and implements a templated implicit cast
> operator for fixed_string<N>, which is where the concatenation is
> actually performed (since it then knows the target type).
> This also works when passing parameters -- at least when those
> parameters are declared with an explicit fixed_string<N> type.
> The caveat is that it doesn't work with "auto" variables, or with
> templated parameters. This limitation unfortunately is quite a large
> one, and can be quite aggravating in practice.
> (Another caveat is that those cases that "don't work" above can easily
> cause dangling references and other lifetime issues instead of cleanly
> failing to compile.)
> Given that, I don't think I'd recommend actually doing it.
> The reference-lifetime-safe alternative is to return an object that
> contains a copy of the parameters, but this is now strictly worse than
> just returning a fixed_string<A+B>. (And, when you then try to assign
> this to a fixed_string<N>, reintroduces the problem of what to do when
> a.size()+b.size() > N, since it will probably often be the case that A+B
> > N, and this is not a mistake.)
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