In the last episode (Oct 10), Li, Qing said:
In the last episode (Oct 08), Li, Qing said:
The bit fields th_x2 and th_off in struct tcphdr, even
though defined as u_int, actually occupies 1 byte.
u_int th_x2:4,/* (unused) */
th_off:4; /* data offset */
The :4 after each variable means 4 bits long, so both fields
together take up 8 bits = 1 byte. That's the whole purpose of
bitfields :)
D'oh
I didn't ask the right question.
It seems u_int specifies the packing and alignment size
for the bit fields, is that correct ?
I don't think so. C99 only allows bitfields to be of type Bool, signed
_int, or unsigned int, so that seems to prevent the use of
char/short/int/long to dictate padding or alignment. There must be
something in the FreeBSD ABI that says structs must be padded so they
are 4-byte aligned, even if none of the members require it. Try putting
your 4 structs into a program and compiling them with gcc -Wpadding:
struct {
u_int a:4,
b:4;
}; is 4 bytes in size.
a.c:7: warning: padding struct size to alignment boundary
struct {
u_int a:4,
b:4;
short c;
}; is 4 bytes in size.
a.c:13: warning: padding struct to align 'c'
(1 byte of padding added just before c)
struct {
u_int a:4,
b:4;
short c;
u_char d;
}; is 8 bytes in size;
a.c:19: warning: padding struct to align 'c'
(1 byte padding just before c, and 3 bytes just after d). I think it
should have printed a padding struct size to alignment boundary
warning also, since if it didn't, the padding after d would have been 1
byte, and struct would have been 6 bytes total.
But
struct {
u_int a:4,
b:4;
u_char d;
short c;
}; is 4 bytes in size;
a.c:21: warning: padding struct size to alignment boundary
This last warning I don't understand, since 1+1+2 is 4 all by itself.
No padding is needed or used.
--
Dan Nelson
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