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jam
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Posted: Wed Nov 29, 2006 8:51 pm Post subject: size of integer types

If I am not mistaken standard says that: sizeof(long)>=sizeof(int)>=sizeof(short)>=sizeof(char)==1 the following lengthes satisfy this condition: 8 > 4 > 2 > 1 ==1 so why int64 on microsoft platform with: sizeof(int64)==8 sizeof(long)==4 sizeof(int)==4 sizeof(short)==2 sizeof(int64)==1 int64 looks ugly regards FM --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ]

Guest

Posted: Wed Nov 29, 2006 10:03 pm Post subject: Re: size of integer types

jam wrote:

Quote:

If I am not mistaken standard says that:

sizeof(long)>=sizeof(int)>=sizeof(short)>=sizeof(char)==1

Actually, it doesn't. The required relationships are in terms of the
range of values represented, not in terms of the amount of memory used
to store the values. It's perfectly legal for sizeof(short) >
sizeof(long), as long as SHORT_MAX < LONG_MAX.

Of course, as a practical matter, just about everyone uses 2's
complement integer types that use every bit they have available, so the
sizeof(type) and the TYPE_MAX macros will usually be in the same order,
for all types.

Quote:

the following lengthes satisfy this condition:
8 > 4 > 2 > 1 ==1
so why int64 on microsoft platform with:
sizeof(int64)==8
sizeof(long)==4
sizeof(int)==4
sizeof(short)==2
sizeof(int64)==1

I presume that's a typo for sizeof(char)==1?

There's a lot of code out there written back in the days when 32-bit
machines were the state of the art, which therefore assumes that long
is 32 bits. I consider that to be defective code - it's always possible
to write code so that it works correctly, regardless of whether long is
32 bits or 64 bits. However, when 64-bit machines came out, as a
practical matter, many implementations chose to deal with such code by
keeping 'long' as a 32-bit type, rather than breaking it by making
'long' a 64-bit type. Instead, they added things like int64 for the new
64 bit type.

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Guest

Posted: Thu Nov 30, 2006 12:05 am Post subject: Re: size of integer types

"Kristof Zelechovski" wrote:

Quote:

Uzytkownik <kuyper (AT) wizard (DOT) net> napisal w wiadomosci
news:1164835979.036922.233990 (AT) j44g2000cwa (DOT) googlegroups.com...
jam wrote:
If I am not mistaken standard says that:

sizeof(long)>=sizeof(int)>=sizeof(short)>=sizeof(char)==1

Actually, it doesn't. The required relationships are in terms of the
range of values represented, not in terms of the amount of memory used
to store the values. It's perfectly legal for sizeof(short)
sizeof(long), as long as SHORT_MAX < LONG_MAX.

Of course, as a practical matter, just about everyone uses 2's
complement integer types that use every bit they have available, so the
sizeof(type) and the TYPE_MAX macros will usually be in the same order,
for all types.

the following lengthes satisfy this condition:
8 > 4 > 2 > 1 ==1
so why int64 on microsoft platform with:
sizeof(int64)==8
sizeof(long)==4
sizeof(int)==4
sizeof(short)==2
sizeof(int64)==1

I presume that's a typo for sizeof(char)==1?

There's a lot of code out there written back in the days when 32-bit
machines were the state of the art, which therefore assumes that long
is 32 bits. I consider that to be defective code - it's always possible
to write code so that it works correctly, regardless of whether long is
32 bits or 64 bits. However, when 64-bit machines came out, as a

The customer need not be happy that your application uses twice as much
memory as it used to.
Although it works correctly, performance suffers.

practical matter, many implementations chose to deal with such code by
keeping 'long' as a 32-bit type, rather than breaking it by making
'long' a 64-bit type. Instead, they added things like int64 for the new
64 bit type.

Long is a 64-bit type in Visual Basic. And let it stay there.

Too late. It's already a 64-bit type on many C/C++ implementations.

Quote:

The name for the 64-bit type was always long long int

Except where it was named 'long', or 'int64' or 'int64_t' or '_int64'
or any of a wide variety of other things. In C99 it has been
standardized as 'int64_t', not 'long long'. C99 has a 'long long' type,
which could be 64 bits, but it could also be 72 bits, or 128 bits, or
any other size greater than 64 bits. These will be included in C++ in
some future version (unless it's already been done - I haven't been
paying as much attention to the TCs as I should).

Quote:

and the rule was that a plain int is equivalent to a short int or to a long
int depending on the platform.

That was commonplace, but not required by any standard, and code which
relied on that was always inherently non-portable.

Quote:

I prefer this convention because I would like the range of long int to be
well defined.

It has never been well-defined, but it has always had well-defined
limits. Adding a requirement that 'int' be the same size as either
'short' or 'long' would not change either of those two facts, it would
just add another constraint on the possible sizes. That constraint
wouldn't prevent 'long' from being 64 bits. 'int' is allowed to be a 64
bit type. For that matter, 'char' is allowed to be a 64 bit type.

Quote:

It is very strange that the basic types are so vaguely defined by the
standard.
It does not happen in any other language I know about.
Sometimes the ranges of variables just must have absolute limits.

Then you'll probably like a feature that was added In the C99 standard,
size-named types. I gather that they will be added to the C++ standard
sometime soon. In C99, they have names like int32_t, for an optional
type with exactly 32 bits, int_fast32_t and int_least32_t, for the
mandatory types with at least 32 bits that are, respectively, fastest
to use and smallest in size.

I believe that for almost all purely internal uses, either the fast or
small types should be used; the exact-sized types should only be used
to meet externally imposed interface requirements. However, I expect
that most people will use exact-sized types routinely, whether or not
they need to.

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Kristof Zelechovski
Guest

Posted: Thu Nov 30, 2006 4:53 am Post subject: Re: size of integer types

Uzytkownik <kuyper (AT) wizard (DOT) net> napisal w wiadomosci
news:1164835979.036922.233990 (AT) j44g2000cwa (DOT) googlegroups.com...

Quote:

jam wrote:
If I am not mistaken standard says that:

sizeof(long)>=sizeof(int)>=sizeof(short)>=sizeof(char)==1

Actually, it doesn't. The required relationships are in terms of the
range of values represented, not in terms of the amount of memory used
to store the values. It's perfectly legal for sizeof(short)
sizeof(long), as long as SHORT_MAX < LONG_MAX.

Of course, as a practical matter, just about everyone uses 2's
complement integer types that use every bit they have available, so the
sizeof(type) and the TYPE_MAX macros will usually be in the same order,
for all types.

the following lengthes satisfy this condition:
8 > 4 > 2 > 1 ==1
so why int64 on microsoft platform with:
sizeof(int64)==8
sizeof(long)==4
sizeof(int)==4
sizeof(short)==2
sizeof(int64)==1

I presume that's a typo for sizeof(char)==1?

There's a lot of code out there written back in the days when 32-bit
machines were the state of the art, which therefore assumes that long
is 32 bits. I consider that to be defective code - it's always possible
to write code so that it works correctly, regardless of whether long is
32 bits or 64 bits. However, when 64-bit machines came out, as a

The customer need not be happy that your application uses twice as much
memory as it used to.
Although it works correctly, performance suffers.

Quote:

practical matter, many implementations chose to deal with such code by
keeping 'long' as a 32-bit type, rather than breaking it by making
'long' a 64-bit type. Instead, they added things like int64 for the new
64 bit type.

Long is a 64-bit type in Visual Basic. And let it stay there.

The name for the 64-bit type was always long long int
and the rule was that a plain int is equivalent to a short int or to a long
int depending on the platform.
I prefer this convention because I would like the range of long int to be
well defined.
It is very strange that the basic types are so vaguely defined by the
standard.
It does not happen in any other language I know about.
Sometimes the ranges of variables just must have absolute limits.
Chris

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Alberto Ganesh Barbati
Guest

Posted: Thu Nov 30, 2006 9:59 am Post subject: Re: size of integer types

kuyper (AT) wizard (DOT) net ha scritto:

Quote:

That's plain wrong. There is an explicit requirement about storage in
3.9.1/2: "There are four signed integer types: â€œsigned charâ€, â€œshort
intâ€, â€œintâ€, and â€œlong int.â€ In this list, each type provides at least
as much storage as those preceding it in the list."

Ganesh

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James Kanze
Guest

Posted: Thu Nov 30, 2006 4:26 pm Post subject: Re: size of integer types

"Kristof Zelechovski" wrote:

Quote:

I'm not sure what you're getting at. If you're concerned with
space, C++ gives you enough alternatives (and you certainly
aren't using long).

Quote:

long is a 64 bit type on most modern machines, I think. But the
standard doesn't guarantee it. It could be 32 bits, or 36, or
48, or 60, or maybe 72 (just to name some of the possibilities
I'm actually familiar with).

Quote:

The name for the 64-bit type was always long long int

Since when? long long was added to C99; it didn't exist before.
And all that is required is that it be at least 64 bits. On
some machines, it might be 72 bits, or 92, or 120.

Quote:

and the rule was that a plain int is equivalent to a short int or to a long
int depending on the platform.

Again, since when. That's likely to be the case where the
machine only has two available sizes. If the machine only has
one available size, and it is at least 32 bits, it's even likely
that short, int and long all have the same size.

Quote:

I prefer this convention because I would like the range of long int to be
well defined.
It is very strange that the basic types are so vaguely defined by the
standard.
It does not happen in any other language I know about.

Really. I think that Java is about the only language which does
define the size. Fortran certainly didn't, nor does Lisp, nor
any of the languages in the Pascal family (including Ada).
Strictly defining the size limits the number of platforms for
which you can have an efficient implementation, and really
doesn't help the programmer that much.

Quote:

Sometimes the ranges of variables just must have absolute limits.

So you impose them. It would be a rather exceptional case where
those limits happened to correspond exactly to the word size of
the machine anyway.

--
James Kanze (GABI Software) email:james.kanze (AT) gmail (DOT) com
Conseils en informatique orientée objet/
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Guest

Posted: Thu Nov 30, 2006 5:24 pm Post subject: Re: size of integer types

Alberto Ganesh Barbati wrote:

Quote:

kuyper (AT) wizard (DOT) net ha scritto:
jam wrote:
If I am not mistaken standard says that:

sizeof(long)>=sizeof(int)>=sizeof(short)>=sizeof(char)==1

Actually, it doesn't. The required relationships are in terms of the
range of values represented, not in terms of the amount of memory used
to store the values. It's perfectly legal for sizeof(short)
sizeof(long), as long as SHORT_MAX < LONG_MAX.

That's plain wrong. There is an explicit requirement about storage in
3.9.1/2: "There are four signed integer types: "signed char", "short
int", "int", and "long int." In this list, each type provides at least
as much storage as those preceding it in the list."

You're right, of course. I'd like to blame my mistake on confusion
between the C and C++ standards, since I monitor newgroups devoted to
both standards. However, I have to admit that I'd never noticed that
this was one of the differences between those two standards. This
should probably be mentioned in appendix C.1.

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Dave Harris
Guest

Posted: Fri Dec 01, 2006 2:12 am Post subject: Re: size of integer types

farid.mehrabi (AT) gmail (DOT) com (jam) wrote (abridged):

Quote:

If I am not mistaken standard says that:

sizeof(long)>=sizeof(int)>=sizeof(short)>=sizeof(char)==1
the following lengthes satisfy this condition:
8 > 4 > 2 > 1 ==1
so why int64 on microsoft platform with:
sizeof(int64)==8
sizeof(long)==4
sizeof(int)==4
sizeof(short)==2
sizeof(int64)==1

int64 looks ugly

Microsoft wanted sizeof(long)==4 for their 64-bit platform to ease porting
from their 32-bit platform. In particular, they wanted binary
compatibility for structures such as:

struct Demo {
long a;
long b;
};

They weren't so concerned about the size of pointers because structures
with pointers are rarely memcpy'd into permanent memory.

I gleaned the above from a blog entry at:
http://blogs.msdn.com/oldnewthing/archive/2005/01/31/363790.aspx

There is much discussion in the comments as to the wisdom of Microsoft's
choice.

-- Dave Harris, Nottingham, UK.

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kwikius
Guest

Posted: Fri Dec 01, 2006 4:08 pm Post subject: Re: size of integer types

James Kanze wrote:

Quote:

Strictly defining the size limits the number of platforms for
which you can have an efficient implementation, and really
doesn't help the programmer that much.

As far as int's go, neither do you know what the safe range of values
is, nor when overflow has occurred. I don't consider that combination
helpful, in fact I consider it downright dangerous. OTOH Anything that
eliminates variability, and increases predictability and reliability is
helpful to me as a programmer and for the repuation of C++ in general
AFAICS.

Cross platform API's generally do provide integer types guaranteeing
e.g 16 ,32 bits. They do this presumably because it helps them too.

regards
Andy Little

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Ron Natalie
Guest

Posted: Fri Dec 01, 2006 9:55 pm Post subject: Re: size of integer types

Kristof Zelechovski wrote:

Quote:

The name for the 64-bit type was always long long int

Huh? C++ doesn't even have that type.

Quote:

and the rule was that a plain int is equivalent to a short int or to a long
int depending on the platform.

Quote:

I prefer this convention because I would like the range of long int to be
well defined.
It is very strange that the basic types are so vaguely defined by the
standard.

The sizes were designed to be implementation defined to accommodate
a variety of machines efficiently. You know that C exists on machines
where sizeof(int) is 36? I've worked on a variety of supercomputers
that have had integer types all over the place.

C does have some typedefs now that assure you of "at least" and
"exactly" a certain number of bits. If you need to rely on this
these are the way to go. Eventually that feature will be subsumed
in to C++.

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James Kanze
Guest

Posted: Fri Dec 01, 2006 11:47 pm Post subject: Re: size of integer types

kwikius wrote:

Quote:

James Kanze wrote:

Strictly defining the size limits the number of platforms for
which you can have an efficient implementation, and really
doesn't help the programmer that much.

As far as int's go, neither do you know what the safe range of values
is, nor when overflow has occurred.

The safe range of values for an int is INT_MIN...INT_MAX. And
of course, you validate your input up front to ensure that your
intermediate values never exceed these ranges. (Often, in fact,
you'll enforce a significantly smaller range.)

Quote:

I don't consider that combination
helpful, in fact I consider it downright dangerous. OTOH Anything that
eliminates variability, and increases predictability and reliability is
helpful to me as a programmer and for the repuation of C++ in general
AFAICS.

There are a very few cases where you might need an exact number
of bits, but most of the time, you either need at least so many
bits (e.g. the day of the month requires 5 bits), or you need
symbolic constants with the maximum and minimum values, so you
can check up front.

Quote:

Cross platform API's generally do provide integer types guaranteeing
e.g 16 ,32 bits. They do this presumably because it helps them too.

Well, you do have to ensure that whatever you write on one
platform has the same format as you read it with on the other.
The representation and the number of bits don't matter, but they
have to be the same at both ends. Thus, for example, cross
platform API's also specify byte order. But we don't want to
specify that in C++, since it would radically penalize some
machines.

--
James Kanze (Gabi Software) email: james.kanze (AT) gmail (DOT) com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
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Keith Thompson
Guest

Posted: Sat Dec 02, 2006 5:32 am Post subject: Re: size of integer types

ron (AT) spamcop (DOT) net (Ron Natalie) writes:
[...]

Quote:

Really? That's at least 288 bits. (I presume you mean
sizeof(int) * CHAR_BIT == 36.)

--
Keith Thompson (The_Other_Keith) kst-u (AT) mib (DOT) org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

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kwikius
Guest

Posted: Sat Dec 02, 2006 6:52 am Post subject: Re: size of integer types

James Kanze wrote:

Quote:

kwikius wrote:
James Kanze wrote:

The safe range of values for an int is INT_MIN...INT_MAX.

A constant is a value that is immutable. Are these macros constants?

regards
Andy Little

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James Kanze
Guest

Posted: Sat Dec 02, 2006 4:53 pm Post subject: Re: size of integer types

kwikius wrote:

Quote:

James Kanze wrote:
kwikius wrote:
James Kanze wrote:

The safe range of values for an int is INT_MIN...INT_MAX.

A constant is a value that is immutable. Are these macros constants?

The are required to evaluate to integral constant expressions,
with the promoted type of the target.

Did you really think that they could change?

--
James Kanze (Gabi Software) email: james.kanze (AT) gmail (DOT) com
Conseils en informatique orientée objet/
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Guest

Posted: Sat Dec 02, 2006 4:53 pm Post subject: Re: size of integer types

kwikius wrote:

Quote:

James Kanze wrote:
kwikius wrote:
James Kanze wrote:

The safe range of values for an int is INT_MIN...INT_MAX.

A constant is a value that is immutable. Are these macros constants?

Yes. Those macros are in the portion of the C++ standard library which
is inherited from the C standard library. The C++ standard does not
describe the C standard library in any detail, deferring to the C
standard for that description. The C standard does require that those
macros expand to constant expressions with implementation-defined
values.

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