small Q on sizeof

Dear Gurus, please refer the following code

typedef struct _iphdr
{
unsigned int h_len:4;
unsigned int version:4;
unsigned char tos;
unsigned short total_len;
} IpHeader;
when I check the size of the above structure,
IpHeader m;
int i = sizeof(m);
system said "8", why? the first two int have taken 8 byte, plus 1 char, plus 2 byte short, it should be 11 byte.
thanx a lot ahead.

Hi,
the value of 8 results explains like this:
your h_len and version struct members are bitfields of four bit each.
They are packed in one byte, wasting three bytes of one integer where the whole bitfield is
arranged in.

typedef struct _iphdr
{
unsigned int h_len:4; //four LSbits, of least sign. byte of integer
unsigned int version:4; //four MSbits, of least sign. byte of integer
unsigned char tos; //two bytes, one byte wasted for "padding" (alignment)
unsigned short total_len;//two bytes
} IpHeader;




Regards
clem

The simple logging/tracing control:
http://www.softsyst.com/activelog_ef.htm

h_len and version use 1 int (4 bytes)
tos takes 1 byte
total_len takes 2 bytes

total: 7 bytes.

BUT: total_len would be at an odd address (odd addresses are inefficient), so a "packing" byte gets inserted to push it to an even address - total 8 bytes.

He who breaks a thing to find out what it is, has left the path of wisdom - Gandalf

Thank you so much for the answer, but according to the logic: the following struct:
typedef struct tagSth{
char ch1;
char ch2;
char ch3;
}
should take 4 byte, for you need padding, but the compiler still report 3 byte. According to my observation, only the odd char used with an even data (such as short) the padding happens. for example the follwing struct size is 4
typedef struct tag sth {
char ch;
short s;
}
thank you for the answer again.

The compiler does that right, no need to align 1 byte types on even addresses only - so you get 3 in the 1. example and the short in the second is then created at an even address, so you get 4.
Don't argue with the compiler :-)



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