Can some one guide me what is word boundary?
google is no good for me for this
thanks in advance
14  19185 
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
Even one of the c++ technical test for a company was asking the same
question.
In fact in Google there are many c++ sites talking about problems with word
boundary.
hope you are convinced.
Regards,
S
ignorance is bliss
"Jack Klein" <ja*******@spam cop.net> wrote in message
news:fm******** *************** *********@4ax.c om... On Sun, 18 Sep 2005 00:15:43 +0100, "Singleton"
<ku*****@homeca ll.co.uk> wrote in comp.lang.c++:
Can some one guide me what is word boundary?
google is no good for me for this
thanks in advance
What made you decide to ask this question here? There is no such term
defined by the C++ language.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html
Hi,
It depends on how large a word is. On a lot of machines today a word is 32
bit. That means on a word boundary is at address
0
4
8
12
16
etc...
ten years ago or so a word used to be 16 (8086) bits and before that 8 bit
was common (6502, 6800, 6809)
Usually there is a bit of a performance gain working with the processors
native size. So using 32 bits and memory addresses that are on a word
boundary can be more efficient. than memory addresses at non-32 bit -16 bit
or even non 32- non- 16 but 8 bit addresses even the latter is smaller you
might end up with slower code when using those sizes and addresses . in your
code.
--
Regards, Ron AF Greve http://moonlit.xs4all.nl
"Singleton" <ku*****@homeca ll.co.uk> wrote in message
news:43******** **@mk-nntp-2.news.uk.tisca li.com... Can some one guide me what is word boundary?
google is no good for me for this
thanks in advance
"Singleton" writes: 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
In that context, a word is the natural/native amount of data fetched from
memory on a single access by a CPU. A sampling of some of the word sizes
over the years (in bits per word): 6, 8, 15, 16, 18, 24, 30, 32, 36, 51, 60.
And a great many more.
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.
A word is supposed to be the "natural" size in which the computer
ordinarily processes data. For most computers these days that would be
32 bits. So a word boundary on most machines would occur every four
bytes in memory and start at a byte address evenly divisible by 4.
Greg
"Greg" <gr****@pacbell .net> wrote in message
news:11******** *************@g 49g2000cwa.goog legroups.com 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.
A word is supposed to be the "natural" size in which the computer
ordinarily processes data. For most computers these days that would be
32 bits. So a word boundary on most machines would occur every four
bytes in memory and start at a byte address evenly divisible by 4.
Greg
On Windows, a WORD is 16 bits on 32 bit machines. A DWORD (double word) is
32 bits.
This may well be because a WORD was 16 bits on earlier 16 bit versions of
Windows and its size has been left at 16 bits for backward compatibility
reasons.
--
John Carson
John Carson wrote: "Greg" <gr****@pacbell .net> wrote in message
news:11******** *************@g 49g2000cwa.goog legroups.com 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.
A word is supposed to be the "natural" size in which the computer
ordinarily processes data. For most computers these days that would be
32 bits. So a word boundary on most machines would occur every four
bytes in memory and start at a byte address evenly divisible by 4.
Greg
On Windows, a WORD is 16 bits on 32 bit machines. A DWORD (double word) is
32 bits.
This may well be because a WORD was 16 bits on earlier 16 bit versions of
Windows and its size has been left at 16 bits for backward compatibility
reasons.
--
John Carson
Yes, but size of the "int" type in C++ is the one that has to match the
size of a machine word. longs, shorts, and especially typedefs may or
may not match.
In fact the reason that WORD and DWORD exist are to insulate C++ source
code from changes in the size of an int. After all, an int is the type
which is the mostly like to change in size. As it did when the platform
moved from 16 to 32 bits, and presumably will again when it moves from
32 to 64 bits. By using typedefs for integer types, the programmer can
be assured of constant sizes. It allows the programmer to be able to
decide when and how to make the transition to the larger types, instead
of having to do it one day after installing the latest C++ compiler.
Greg
"Greg" <gr****@pacbell .net> wrote in message
news:11******** **************@ g43g2000cwa.goo glegroups.com John Carson wrote: "Greg" <gr****@pacbell .net> wrote in message
news:11******** *************@g 49g2000cwa.goog legroups.com 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.
A word is supposed to be the "natural" size in which the computer
ordinarily processes data. For most computers these days that would
be 32 bits. So a word boundary on most machines would occur every
four bytes in memory and start at a byte address evenly divisible
by 4.
Greg
On Windows, a WORD is 16 bits on 32 bit machines. A DWORD (double
word) is 32 bits.
This may well be because a WORD was 16 bits on earlier 16 bit
versions of Windows and its size has been left at 16 bits for
backward compatibility reasons.
--
John Carson
Yes, but size of the "int" type in C++ is the one that has to match
the size of a machine word. longs, shorts, and especially typedefs
may or may not match.
In fact the reason that WORD and DWORD exist are to insulate C++
source code from changes in the size of an int. After all, an int is
the type which is the mostly like to change in size. As it did when
the platform moved from 16 to 32 bits, and presumably will again when
it moves from 32 to 64 bits. By using typedefs for integer types, the
programmer can be assured of constant sizes. It allows the programmer
to be able to decide when and how to make the transition to the
larger types, instead of having to do it one day after installing the
latest C++ compiler.
Greg
I agree with most of what you say, but in the interests of accuracy would
point out:
1. The distinction that you wish to make between word (lowercase) and the
typedef WORD (uppercase) is not one that is maintained in Windows
programming. I am not defending Windows terminology, just pointing it out.
See here for example: http://msdn.microsoft.com/library/de...ros/loword.asp
2. 64 bit Windows is already here and an int is still 32 bits (a
controversial decision that could easily have gone the other way, but there
you have it).
--
John Carson
"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
Choice 2 : On the stack
Choice 3 : Initialized
Choice 4 : On a word boundary
Choice 5 : On the heap
The correct answer is Choice 6: None of the above.
Objects with virtual base classes are not always contiguous, which rules out
(1).
The word "stack" appears in the C++ standard only in the context of "stack
unwinding" that occurs as part of exception handling, and in the
descriptions of the library algorithms that deal with stacks.
Objects are not guaranteed to be initialized.
The phrase "word boundary" appears nowhere in the C++ standard.
The word "heap" appears in the C++ standard only in the context of
describing the library algorithms that deal with heaps. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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