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Subject: Re: [Boost-users] Boost-users Digest, Vol 1983, Issue 1
From: John Dlugosz (JDlugosz_at_[hidden])
Date: 2009-05-04 12:29:10

> Date: Fri, 01 May 2009 20:15:50 -0600
> From: tom fogal <tfogal_at_[hidden]>
> Subject: Re: [Boost-users] General C++: Class Method Signatures
> Miss-Match
> To: boost-users_at_[hidden]
> Message-ID: <auto-000019286349_at_[hidden]>
> Dominique Devienne <ddevienne_at_[hidden]> writes:
> > On Fri, May 1, 2009 at 10:32 AM, Dan Day <coolmandan_at_[hidden]>
> wrote:
> > > I thought that int represented the natural word size of the
> > > architecture. If so, then wouldn't unsigned int be 64-bit on a
> > > 64-bit platform as well?
> I had heard this before as well. I took a quick look through c1x-1336
> (current (?) draft of C) though, and I could not find anything to this
> effect.
> Of course, that doesn't say it's not there, and I did not check the
> standard.
> > The few 64-bit platforms I've seen keep [unsigned] int at
> > 32-bits. --DD

What is "natural"? The AMD-x64 is built on top of the 32-bit
architecture. The instruction set (in 64-bit mode) uses 32 bits as the
nominal and default data size for most operations. Even for memory
addresses, it's a 32-bit relative-to-IP value.

It takes an opcode prefix to indicate the use of 64-bit values.
The memory architecture can handle 32-bit values without penalty, and
you fit more of them in the cache. So 32-bit ints are not at all

In principle, using 32-bit ints for casual use instead of gratuitously
using 64-bits where you don't need the extra size will run faster. It
reduces the memory bandwidth for both data and code size, even though
the internal execution is the same speed in either case.

(sorry about the footer; it's not my idea)

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