what is word boundary?

[Singleton]

Can some one guide me what is word boundary?
google is no good for me for this
thanks in advance

[Jack Klein]

On Sun, 18 Sep 2005 00:42:33 +0100, "Singleton"
<ku*****@homeca ll.co.uk> wrote in comp.lang.c++:

I just took c++ exam in brainbnch . one of the question is
Q:)According to the C++ standard, what is an object's internal
representation in memory guaranteed to be?
Choice 1 : Contiguous
Choice 2 : On the stack
Choice 3 : Initialized
Choice 4 : On a word boundary
Choice 5 : On the heap

None of these answers is correct according to the C++ standard. And
C++, as I said, does not even define the term "word boundary". This
merely means that Brainbench has made a mistake in their test. In the
past, they have worked to correct such errors when pointed out to
them. I do not know whether they still do so or not.

--
Jack Klein
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[Jack Klein]

On 17 Sep 2005 18:26:25 -0700, "Greg" <gr****@pacbell .net> wrote in
comp.lang.c++:

Singleton wrote:

Can some one guide me what is word boundary?
google is no good for me for this
thanks in advance

The size of an int in a C++ compiler should be the same as the size of
a word for the architecture that that compiler targets.

Yes, that is a an intention expressed by the C language standard, and
inherited by C++. But there are C and even some C++ compilers for
architectures that are strictly 8-bit, and have no 16-bit registers at
all, yet here int must still be at least 16 bits.

--
Jack Klein
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[Old Wolf]

Andrew Koenig wrote:

"Singleton" <ku*****@homeca ll.co.uk> wrote in message
news:43******** **@mk-nntp-2.news.uk.tisca li.com...

I just took c++ exam in brainbnch . one of the question is

Q:)According to the C++ standard, what is an object's internal
representation in memory guaranteed to be?
Choice 1 : Contiguous

Objects with virtual base classes are not always contiguous,
which rules out (1).

The C++ standard requires that the memory for arrays (and vectors)
must be contiguous. This seems to imply that the objects stored
in the array or vector must also use contiguous memory.

What's your definition of 'contiguous' ?

[EventHelix.com]

The word boundary refers to word level aligment in memory. If the word
is 32 bit, it means that words are aligned at addresses that are
divisible by 4.

The following article should help:
http://www.eventhelix.com/RealtimeMa...ndOrdering.htm

--
EventStudio 2.5 - http://www.EventHelix.com/EventStudio
Sequence Diagram Based System Design and Modeling Tool

[Andrew Koenig]

"Old Wolf" <ol*****@inspir e.net.nz> wrote in message
news:11******** ************@g1 4g2000cwa.googl egroups.com...

Objects with virtual base classes are not always contiguous,
which rules out (1).

The C++ standard requires that the memory for arrays (and vectors)
must be contiguous. This seems to imply that the objects stored
in the array or vector must also use contiguous memory. What's your definition of 'contiguous' ?

Good question. Consider this:

struct A { int a; };
struct B : public virtual A { int b; };
struct C : public virtual A { int c; };
struct D: public virtual B, public virtual C { ind d; };

D d;
B* bp = &d;
C* cp = &d;

Now bp points to an object of type B and cp points to an object of type C.
Each of these objects has a member named a, and because both objects are
subobjects of the same object (namely d), bp->a and cp->a are the same
object. In other words, &bp->a == &cp->a.

The pointer bp points to an object (*bp) of type B, which has two members: b
(defined directly as part of the type) and a (inherited from the base class
A). I think that at the very least, if you say that *bp is contiguous, you
are saying that no other objects appear between bp->a and bp->b. They might
not occupy adjacent addresses, but there can't be any other objects between
them.

Therefore, if *bp is contiguous, bp->a and bp->b are in adjacent memory with
the possible exception of padding or other compiler-generated bookkeeping.

So is *bp contiguous? Well, the answer is either yes or no. If it's no,
then there is such a thing as a discontiguous object. So let's assume that
the answer is yes and forge on. This yes answer implies that bp->a and
bp->b have no other objects between them.

Now let's look at *cp, the object to which cp points. It has two members,
cp->a and cp->c. Is *cp contiguous?

If *cp is contiguous, cp->a and cp->c must have no other objects between
them. However, cp->a is the same object as bp->a, and we have already
established that there is an object (namely bp->b) adjacent to bp->a.
Therefore cp->c cannot be adjacent to cp->a, and *cp cannot be contiguous!

In other words, I have proven that after executing the code shown above, it
is impossible for bp and cp both to point to contiguous objects, according
to a fairly loose definition of contiguity.