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Subject: Re: [boost] gcc4.3 warning for multiple scope exits
From: Alexander Nasonov (alnsn_at_[hidden])
Date: 2009-07-14 09:38:56
Kim Barrett <kab.conundrums <at> verizon.net> writes:
>
> At 10:48 AM +0000 7/14/09, Alexander Nasonov wrote:
> >boost::scope_exit::aux::holder<
> > boost::scope_exit::aux::declare<
> > sizeof(boost_scope_exit_args)
> > >::apply<0>::apply_value
> >> boost_scope_exit_args;
> >
> >If sizeof(boost_scope_exit_args) is equal to sizeof(declared),
> >the resulting type would be holder<(0<0>::apply_value)>. But I don't
> >see a global variable apply_value in your patch.
>
> The added parenthesis in the above, in
>
> holder<(0<0>::apply_value)>
>
> don't describe the actual parse and evaluation. Instead, that snippet
> gets parsed and evaluated as
>
> holder<(0<0)>::apply_value
>
This is crazy, isn't it?
Q: correctly indent the following code: <see code above>
A: it's impossible
For the second use, the correct indentation is:
boost::scope_exit::aux::holder<
boost::scope_exit::aux::declare<
sizeof(boost_scope_exit_args)
>::apply < 0
>::apply_value > boost_scope_exit_args;
> >BTW, I modified Steven's code and introduced a bug because
> >sizeof(declare<sizeof(undeclared)>::apply<0>) does not necessarily
> >equal to sizeof(declared).
>
> Are you sure? Perhaps I'm missing something, but it looks to me like
> those values should indeed be the same. The "apply" class template in
> question has one member, of type "declared". Given that, I would be
> surprised if the "apply" class template could have a different size
> (or alignment) from the "declared" type.
They're the same because void* is well aligned and there is no padding
in struct apply<0> but strictly speaking, they don't have to be of equal
size. I like explicit specializations for declared and undeclared, though.
They make reading code a little bit easier (and here "a little bit" make
a big difference, given how complex the trick is).
Perhaps adding BOOST_STATIC_ASSERT would be enough?
Alex
multiplied by a complec.
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