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Subject: Re: [boost] [metaparse] performance comparisons?
From: Louis Dionne (ldionne.2_at_[hidden])
Date: 2015-06-02 11:09:41


Peter Dimov <lists <at> pdimov.com> writes:

>
> Louis Dionne wrote:
> > > I assume that you mean for 'f' to be constexpr?
> >
> > No, it does not change a thing. It was on purpose that I did _not_ make it
> > constexpr, in order not to mix concerns. What we're interested in is the
> > constexpr-ness of `s` within the body of `f`, whether `f` is constexpr or
> > not does not change anything.
>
> I know that it doesn't, but we're demonstrating that 's' isn't constexpr
> even when 'f' is, which is a stronger claim than that 's' isn't constexpr
> when 'f' isn't.

Sure, I agree with your point. However, just to clarify my view of things,
I see f's constexpr-ness as being orthogonal (i.e. unrelated) to the
constexpr-ness of its argument. Hence, I do not really perceive it as a
stronger claim, just a different one that might make things trickier
for someone not understanding this orthogonality.

But I guess this boils down to pedagogy, and I think we have no real
technical disagreement here.

Regards,
Louis


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