On Wed, Jul 30, 2003 at 02:28:41AM -0700, Muhammad Farooq-i-Azam wrote:
> I think I can define a structure as below for the
> fixed part of the header:
>
> typedef struct header_ripv1 {
> u_int8_t command;
> u_int8_t version;
> u_int16_t zeros;
> } header_ripv1_t;
>
> My question is can I use the above structure for both
> little endian and big endian systems with the only
> conversion required below:
>
> header_ripv1_t ripv1;
>
> ripv1.zeros = ntohs(ripv1.zeros);
Yes.
> Or do I have to do something like this (which I
> *doubt* has a double error, and will result in
> problems on a big endian system):
>
> typedef struct header_ripv1{
> #if defined (WORDS_BIGENDIAN)
> u_int16_t zeros;
> u_int8_t version;
> u_int8_t command;
> #else
> u_int8_t command;
> u_int8_t version;
> u_int16_t zeros;
> #endif
> } header_ripv1_t;
No. The byte-order issue is only an issue of the order of the bytes
within a multi-byte numerical quantity; it isn't an issue of the order
of the bytes within a structure. ANSI C, as I remember, specifies that,
with a structure like
typedef struct header_ripv1 {
u_int8_t command;
u_int8_t version;
u_int16_t zeros;
} header_ripv1_t;
"command" is in the first location, followed by "version", followed by
"zeroes".
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