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dcorbit@connx.com
*nix forums addict
Joined: 13 Jun 2006
Posts: 92
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 Posted: Fri Jul 21, 2006 12:21 am Post subject:
Re: sizeof()
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"Ronald Bruck" <bruck@math.usc.edu> wrote in message
news:200720061659215633%bruck@math.usc.edu...
[snip]
| Quote: | This brings up a problem I've wondered about: what support is there in
C for two-byte characters? I can foresee, in not too many years, the
ordinary char going the way of the "short"--provided for efficiency,
but not often used. That way, I'll be able to get even MORE junk
e-mail in Kanji :-(
Googling and Wikipedia-ing, I find info on XDR (eXternal Data
Representation), but nothing specifically about C, except a library
provided by IBM. Is this done externally to C? Is there some standard
library for it?
|
You are referring to ICU, I imagine.
http://www-306.ibm.com/software/globalization/icu/index.jsp
That is the only tool set that I know of that is full featured enough to
solve the typical problem set associated with internationalization.
If you need to do internationalization of your software projects, then I can
say that I highly recommend it. We use it with C++ but I gather that it can
also be used from C. |
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Ronald Bruck
*nix forums beginner
Joined: 13 Mar 2006
Posts: 11
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| Posted: Thu Jul 20, 2006 11:59 pm Post subject:
Re: sizeof()
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In article <uh40c255petibtrmh91023l3ksq8ghpm29@4ax.com>, Barry Schwarz
<schwarzb@doezl.net> wrote:
| Quote: | On Thu, 20 Jul 2006 03:33:18 GMT, dbtid <dbtid0@nospam.gmail.com
wrote:
Tom St Denis wrote:
Jack wrote:
For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
why not?
Tom
What a useless answer.
Why not just tell him there is padding?
Because sizeof(int) could be 7, sizeof(double) could be 8, and
sizeof(char) must be 1. No padding at all.
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This brings up a problem I've wondered about: what support is there in
C for two-byte characters? I can foresee, in not too many years, the
ordinary char going the way of the "short"--provided for efficiency,
but not often used. That way, I'll be able to get even MORE junk
e-mail in Kanji :-(
Googling and Wikipedia-ing, I find info on XDR (eXternal Data
Representation), but nothing specifically about C, except a library
provided by IBM. Is this done externally to C? Is there some standard
library for it?
--
Ron Bruck
Posted Via Usenet.com Premium Usenet Newsgroup Services
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
----------------------------------------------------------
http://www.usenet.com |
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Barry Schwarz
*nix forums Guru
Joined: 09 Apr 2005
Posts: 390
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| Posted: Thu Jul 20, 2006 11:33 pm Post subject:
Re: sizeof()
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On Thu, 20 Jul 2006 03:33:18 GMT, dbtid <dbtid0@nospam.gmail.com>
wrote:
| Quote: | Tom St Denis wrote:
Jack wrote:
For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
why not?
Tom
What a useless answer.
Why not just tell him there is padding?
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Because sizeof(int) could be 7, sizeof(double) could be 8, and
sizeof(char) must be 1. No padding at all.
Remove del for email |
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Giorgio Silvestri
*nix forums beginner
Joined: 08 Jul 2006
Posts: 4
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| Posted: Thu Jul 20, 2006 5:10 pm Post subject:
Re: sizeof()
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"Richard Tobin" <richard@cogsci.ed.ac.uk> ha scritto nel messaggio
news:e9o2q3$18ar$1@pc-news.cogsci.ed.ac.uk...
| Quote: | In article <1153403331.267938.304730@s13g2000cwa.googlegroups.com>,
sarathy <sps.sarathy@gmail.com> wrote:
and you would be asking the system to read an int from an odd byte
address. A lot of systems cannot do that
What does it mean that system cannot read odd byte address ?
He said read *an int* from an odd byte address. Many modern machines
can read a single byte from any location, but a 2-byte word only from
even addresses, a 4-byte word only from multiple-of-4 addresses, and
so on.
Consider that word length is 4. For the struct,
struct a
{
char c;
};
int main()
{
struct a b[10];
printf ("size = %d\n",sizeof(b));
}
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No.
sizeof(b) is of type 'size_t'.
It can be bigger than 'int'.
#include <stdio.h>
#include <stdlib.h>
struct a
{
char c;
};
int main(void)
{
struct a b[10];
/* I suppose sizeof(b) <= INT_MAX  */
printf ("size = %d\n",(int)sizeof(b));
return EXIT_SUCCESS;
}
In C99 you can use also %zu.
Giorgio Silvestri |
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Richard Tobin
*nix forums Guru
Joined: 22 Mar 2005
Posts: 372
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| Posted: Thu Jul 20, 2006 2:09 pm Post subject:
Re: sizeof()
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In article <1153403331.267938.304730@s13g2000cwa.googlegroups.com>,
sarathy <sps.sarathy@gmail.com> wrote:
| Quote: | and you would be asking the system to read an int from an odd byte
address. A lot of systems cannot do that
What does it mean that system cannot read odd byte address ?
|
He said read *an int* from an odd byte address. Many modern machines
can read a single byte from any location, but a 2-byte word only from
even addresses, a 4-byte word only from multiple-of-4 addresses, and
so on.
| Quote: | Consider that word length is 4. For the struct,
struct a
{
char c;
};
int main()
{
struct a b[10];
printf ("size = %d\n",sizeof(b));
}
As per the discussion, should the answer be 40 (3 bytes padded for
evrey 1 char)????
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In this case, you never need to read anything bigger than char, so alignment
isn't likely to be a problem. sizeof(b) will probably be 10.
-- Richard |
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sarathy
*nix forums beginner
Joined: 17 Jul 2006
Posts: 12
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| Posted: Thu Jul 20, 2006 1:48 pm Post subject:
Re: sizeof()
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Walter Roberson wrote:
| Quote: | In article <1153324264.506809.63940@i3g2000cwc.googlegroups.com>,
Jack <junw2000@gmail.com> wrote:
For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
That implementation probably uses 4 bytes for int, 8 bytes for double,
1 byte for char. But implementations are required to pad structures so
that the size of the structure is a multiple of the architectural
alignment restrictions (often 4 bytes), so the structure likely gets
padded by adding 3 unused bytes.
If the structure were not padded, then if you had
struct A sa2[2];
then the address of sa2[1] would be 13 bytes after the start of sa2[0]
and you would be asking the system to read an int from an odd byte
address. A lot of systems cannot do that, or can only do that by using
special trap handlers, or can only do it by using special
"load unaligned" instructions that are often very slow.
--
"law -- it's a commodity"
-- Andrew Ryan (The Globe and Mail, 2005/11/26)
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What does it mean that system cannot read odd byte address ?
Is this due to the concept of word length, which is the basic length of
a memory address.
If such is the case, then how is memory allocated for following case.
Consider that word length is 4. For the struct,
struct a
{
char c;
};
int main()
{
struct a b[10];
printf ("size = %d\n",sizeof(b));
}
As per the discussion, should the answer be 40 (3 bytes padded for
evrey 1 char)????
so that there is not need for reading odd byte address?
Correct me if this not true.
Sarathy |
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jaysome
*nix forums addict
Joined: 01 Mar 2006
Posts: 69
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| Posted: Thu Jul 20, 2006 8:40 am Post subject:
Re: sizeof()
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On 19 Jul 2006 08:58:01 -0700, "Robert Gamble" <rgamble99@gmail.com>
wrote:
| Quote: | Jack wrote:
For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
Read the FAQ, <http://c-faq.com/>, question 2.13.
|
which says:
Q. Why does sizeof report a larger size than I expect for a structure
type, as if there were padding at the end?
A. Padding at the end of a structure may be necessary to preserve
alignment when an array of contiguous structures is allocated. Even
when the structure is not part of an array, the padding remains, so
that sizeof can always return a consistent size. See also question
2.12.
References: H&S Sec. 5.6.7 pp. 139-40
Best regards
--
jay
NOTE: The C FAQ book has the same contents as the online C FAQ and
more. The online version is great and the book is even better. Find
out more here:
http://c-faq.com/book/ |
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dbtid
*nix forums beginner
Joined: 03 Jul 2006
Posts: 3
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| Posted: Thu Jul 20, 2006 3:33 am Post subject:
Re: sizeof()
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Tom St Denis wrote:
| Quote: | Jack wrote:
For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
why not?
Tom
|
What a useless answer.
Why not just tell him there is padding? |
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Roberto Waltman
*nix forums Guru Wannabe
Joined: 21 Feb 2005
Posts: 180
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| Posted: Wed Jul 19, 2006 4:06 pm Post subject:
Re: sizeof()
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"Jack" <junw2000@gmail.com> wrote:
| Quote: | For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
|
Probably because sizeof(int) is 4, sizeof(double) is 8, sizeof(char)
is 1 (of course,) and 3 chars of padding are added to guarantee proper
alignment.
But it could also be that in your machine sizeof(int) is 1,
sizeof(double) is 14 and no padding is necessary.
Or maybe sizeof(int) is 7, followed by one char padding,
sizeof(double) is 5, followed by one char padding, plus the size of c
and one last char of padding for alignment.
Or any other valid combination adding up to 16.
Check your compiler's documentation for the sizes of the primitive
data types and alignment requirements.
Roberto Waltman
[ Please reply to the group,
return address is invalid ] |
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Walter Roberson
*nix forums Guru
Joined: 19 Feb 2005
Posts: 1300
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| Posted: Wed Jul 19, 2006 4:01 pm Post subject:
Re: sizeof()
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In article <1153324264.506809.63940@i3g2000cwc.googlegroups.com>,
Jack <junw2000@gmail.com> wrote:
| Quote: | For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
|
That implementation probably uses 4 bytes for int, 8 bytes for double,
1 byte for char. But implementations are required to pad structures so
that the size of the structure is a multiple of the architectural
alignment restrictions (often 4 bytes), so the structure likely gets
padded by adding 3 unused bytes.
If the structure were not padded, then if you had
struct A sa2[2];
then the address of sa2[1] would be 13 bytes after the start of sa2[0]
and you would be asking the system to read an int from an odd byte
address. A lot of systems cannot do that, or can only do that by using
special trap handlers, or can only do it by using special
"load unaligned" instructions that are often very slow.
--
"law -- it's a commodity"
-- Andrew Ryan (The Globe and Mail, 2005/11/26) |
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Robert Gamble
*nix forums Guru
Joined: 15 Apr 2005
Posts: 447
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| Posted: Wed Jul 19, 2006 3:58 pm Post subject:
Re: sizeof()
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Jack wrote:
| Quote: | For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
|
Read the FAQ, <http://c-faq.com/>, question 2.13.
Robert Gamble |
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Tom St Denis
*nix forums Guru Wannabe
Joined: 22 Feb 2005
Posts: 161
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| Posted: Wed Jul 19, 2006 3:54 pm Post subject:
Re: sizeof()
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Jack wrote:
| Quote: | For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
|
why not?
Tom |
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junw2000@gmail.com
*nix forums addict
Joined: 24 Feb 2006
Posts: 92
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| Posted: Wed Jul 19, 2006 3:51 pm Post subject:
sizeof()
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For the structure below:
struct A{
int a;
double b;
char c;
};
struct A sa;
printf("sizeof(A): %d\n", sizeof(sa));
The output is
sizeof(A): 16
Why it is 16?
Thanks.
Jack |
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