Re: using sizeof
Jon Dowland wrote:
Because sizeof is not really the size of the struct, it is
the distance between adjacent structs in an array.
Alignment forces the extra bytes
I'm not quite sure I get what you're saying here. Yes,
alignment pads out the structure. But I'm not sure where
arrays come into it :- sizeof(struct whatever) is applicable
to a single instance of a struct too.
What he's saying is that
struct FredStruct Fred[2];
...
sizeof(struct Fred[0])
is the same as
(void *)(Fred[1])-(void *)(Fred[0])
Structures may get padding at the end precisely for this
reason. It is not obvious that this is the case when one
talks about a single instance.
Mike
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Re: using sizeof
On Mon, 14 May 2007 16:24:59 +, J HU wrote: Dear all, Perhaps it's an easy/silly question but I don't understand how it works. (I'm working in a debian) I have declared a structure and I'm using the sizeof to get the size of this structure. After the call I get that the total size is 64Bytes but if I get the size of each field and I add them manually I get that it should be 61Bytes... Anyone knows why the result is not the same? Thanks in advance Javi Because sizeof is not really the size of the struct, it is the distance between adjacent structs in an array. Alignment forces the extra bytes. In my opinion, 'sizeof' should have been called 'spacingof', and there should have been other properties for alignment, such as sizeof, modulusof. and remainderof, so you could do your own storage allocation calculations in a machine-independent way. Something could take sizeof bytes starting at an address that is remainderof bytes past a n address that is a multiple of modulusof. The log2 of modulus would also be useful. -- hendrik -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
Re: Re: using sizeof
Because sizeof is not really the size of the struct, it is the distance between adjacent structs in an array. Alignment forces the extra bytes I'm not quite sure I get what you're saying here. Yes, alignment pads out the structure. But I'm not sure where arrays come into it :- sizeof(struct whatever) is applicable to a single instance of a struct too. -- Jon Dowland -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
using sizeof
Dear all,
Perhaps it's an easy/silly question but I don't understand how it works.
(I'm working in a debian)
I have declared a structure and I'm using the sizeof to get the size of
this structure.
After the call I get that the total size is 64Bytes but if I get the size of
each field and I add them manually I get that it should be 61Bytes...
Anyone knows why the result is not the same?
Thanks in advance
Javi
Code:
#include stdio.h
#include sys/socket.h
typedef struct Packet{
//char transportIdentifier;
int mode;
int idPacket;
int idMessage;
int fragmentNumber;
struct sockaddr source;
struct sockaddr destination;
int length;
int totalLength;
int* groupSet;
unsigned char data[1];
};
int main (){
Packet infoPacket;
printf(Size of mode %d\n,sizeof(infoPacket.mode));
printf(Size of idPacket %d\n,sizeof(infoPacket.idPacket));
printf(Size of idMessage %d\n,sizeof(infoPacket.idMessage));
printf(Size of fragmentNumber %d\n,sizeof(infoPacket.fragmentNumber));
printf(Size of source %d\n,sizeof(infoPacket.source));
printf(Size of destination %d\n,sizeof(infoPacket.destination));
printf(Size of length %d\n,sizeof(infoPacket.length));
printf(Size of totalLength %d\n,sizeof(infoPacket.totalLength));
printf(Size of groupSet %d\n,sizeof(infoPacket.groupSet));
printf(Size of data %d\n,sizeof(infoPacket.data));
printf(Total Size of infoPacket = %d or %d or
%d\n,sizeof(infoPacket),sizeof(Packet),sizeof(struct Packet));
}
Output:
Size of mode 4
Size of idPacket 4
Size of idMessage 4
Size of fragmentNumber 4
Size of source 16
Size of destination 16
Size of length 4
Size of totalLength 4
Size of groupSet 4
Size of data 1
Total Size of infoPacket = 64 or 64 or 64
_
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Re: using sizeof
J HU [EMAIL PROTECTED] wrote:
I have declared a structure and I'm using the sizeof to get the size of
this structure.
After the call I get that the total size is 64Bytes but if I get the size
of each field and I add them manually I get that it should be 61Bytes...
Anyone knows why the result is not the same?
61 bytes doesn't fit evenly on a 32-bit boundary; 64 bytes does. Your
struct is padded so that your processor can page through them faster.
If you really want your struct to take up only the 61 bytes for it's
structure, you can use the packed attribute;
struct foo {
int a;
int b;
int c;
} __attribute__((__packed__));
Kinda odd syntax, but it works. :) Really, it's a lot faster for your
processor to deal with structs that start on even 32-bit boundaries, so
only pack the structure if you absolutely need it (eg; to speak a particular
networking/storage protocol...)
- Tyler
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Re: using sizeof
On 5/14/07, J HU [EMAIL PROTECTED] wrote: Dear all, Perhaps it's an easy/silly question but I don't understand how it works. (I'm working in a debian) I have declared a structure and I'm using the sizeof to get the size of this structure. After the call I get that the total size is 64Bytes but if I get the size of each field and I add them manually I get that it should be 61Bytes... Anyone knows why the result is not the same? I noticed you got an answer. This page can help also: http://en.wikipedia.org/wiki/Data_structure_alignment The effects of alignment in performance (and even stability, if you're writing low level code) depend on the architecture. For instance: A memory access can be much faster on some architectures if the base address for the request is aligned. So, if you can afford the padding space in memory, also remember that you have to be careful with pointer arithmetic. You should not assume that: *((char*)(pointer + sizeof(mystruct) - 1)) Is the last byte of your structure. Regards, Nelson.- -- http://arhuaco.org http://emQbit.com -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
