Subject: Re: [boost] [boost::endian] Request for comments/interest
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2010-06-03 11:17:36
Terry Golubiewski wrote:
> Robert Stewart wrote:
> > Terry Golubiewski wrote:
> > (Since buf is a void * in the snipped code, static_cast
> > would work for your casts, too.)
> 'buf' must be a pointer to "something". The void* is just
> masking what is really analagous to a reinterpret_cast.
> void* is just as ugly as reinterpret_cast. Maybe more so.
Data read from disk is untyped. Many APIs use void *, others char *. Either way, there's no useful type information. If you have a void *, you can use static_cast to another pointer type. If you have a char *, you must use reinterpret_cast to another pointer type (or two static_casts).
> >> What bugs me is that 'buf' doesn't point to a 'T' yet. It
> >> points to a half-baked 'T'. Only after one calls
> >> swap_in_place() is the object really a 'T'.
> > That would apply to floating point, certainly. However,
> > for integer types, your statement is false. Before the
> > swap, they are still integers, they just don't have the
> > desired value.
> I disagree. Physically, they are compatible, but logically, they are
We're in violent agreement.
> I think the logical view matters more.
I can't argue that point, but I don't think it signifies. In either case, the data read from disk is not usable until endianness has been addressed. Once it has, all remaining logic works with the appropriate type.
> > Even in the object-based approach, the object isn't a
> > proper T until you access it and the swap occurs. The
> > implicit swap veneer just gives you the impression
> > that it is already in the desired form.
> An endian<X, T> is not the same as a T. Even
> endian<native,T> must be converted to a T before you can
> operate on it. The veneer is important, because it
> works like a T, without actually being a T. I think
> that subtle distinction is important and is the primary
> motivation for type-based endian.
You've made a reasonable case. "Object-based" please.
> I object to your use of the word "swap" though. My
> type-based approach does a reverse_copy, not a swap.
> This distinction is the fundamental topic of this
Perhaps you posted code that I was expected to have perused to know that your code did a reverse copy rather than a swap, but I didn't see such code or notice that it did so. However, a reverse copy is not always correct for changing endianness, though certainly so for changes between big and little endianness.
Rob Stewart robert.stewart_at_[hidden]
Software Engineer, Core Software using std::disclaimer;
Susquehanna International Group, LLP http://www.sig.com
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