Re: [hash2][review] Early review (due to holidays)

From: Claudio DeSouza (cdesouza_at_[hidden])
Date: 2024-12-06 18:46:56

detail::read32le( block.subspan(i * 4).first<4>());

So you also get a sized span, and avoid checks.

On Fri, 6 Dec 2024 at 17:42, Andrey Semashev via Boost <
boost_at_[hidden]> wrote:

> On 12/6/24 20:30, Peter Dimov via Boost wrote:
> > You can see practical manifestations of these examples in e.g
> >
> > https://github.com/pdimov/hash2/blob/develop/include/boost/hash2/md5.hpp
> >
> > where at
> >
> >
> https://github.com/pdimov/hash2/blob/26b16023dc639ae354e246458719e4b569bc1d97/include/boost/hash2/md5.hpp#L110
> >
> > you have
> >
> >     BOOST_CXX14_CONSTEXPR void transform( unsigned char const block[ 64
> ] )
> >
> > with the implicit precondition that `block` points at at least 64 valid
> bytes,
> > and then at
> >
> >
> https://github.com/pdimov/hash2/blob/26b16023dc639ae354e246458719e4b569bc1d97/include/boost/hash2/md5.hpp#L119-L122
> >
> > you have
> >
> >         for( int i = 0; i < 16; ++i )
> >         {
> >             x[ i ] = detail::read32le( block + i * 4 );
> >         }
> >
> > which relies on this implicit precondition to read 16 subspans of size 4.
>
> Assuming you had code like this:
>
>   void transform( span<unsigned char const, 64> block )
>   {
>     for( int i = 0; i < 16; ++i )
>     {
>       x[ i ] = detail::read32le( block.subspan(i * 4, 4) );
>     }
>   }
>
> I would presume, the subspan call would have a runtime check for the
> offset and size. Then the converting constructor from span<const
> unsigned char> returned from subspan to span<const unsigned char, 4>
> accepted by read32le would have to check the size again. Yes, the
> compiler could optimize those checks away, if it's smart enough. Or it
> could not, e.g. if optimizations are disabled or it fails to propagate
> constants for some reason. But none of these checks are actually
> necessary, except maybe one at the entry, to enforce the size of the
> entire block. Maybe not even that, if the callers of transform are
> guaranteed to always pass the whole blocks.
>
>
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