Hello there to all :)
Maybe this could be a newbie question, but I never though at
this scenario before yestarday and I cannot try it to be sure,
so ... I'm asking here :)
Let's consider this normal little data transfer between an
alpha architecture machine and an i386 one.
Host A will be the machine running on top of an alpha processor.
Host I will be the machine running on top of an i386 processor.
(Nontheless to say that both A and I are running Linux as OS)
Host A:
[..]
unsigned long data = 0xffffffffffffffff;
data = htonl(data);
send(fd, (char *) &data, sizeof(data), 0);
where, obviously `fd' represent a connected socket to the `I' machine.
(We're using TCP as Transport Protocol)
Host I:
unsigned long data;
recv(fd, (char *)&data, sizeof(data), 0);
data = ntohl(data);
printf("data received: %ld\n", data);
where here too, fd represent a connected socket (TCP) to `A' machine.
On both machine, ntohs/htons will handle the right byte ordering stuff.
TCP and IP will transmit the right number of byte (8 on alpha for a long)
What will happen at receive time on the host `I' (i386) where a long is
4 byte long ?
Shall I see 0xffffffffffffffff or shall I lost some info ?
Well, the info isn't lost. It will remain in the TCP input buffer queue
`till a further recv(). Right ?
But with only a recv() I'll get only 4 byte (as sizeof(long) on i386)
Is all this right ?
Could someone give me other infos ?
Thx a lot and ... Have a nice day/night ... well whatever :))
bye bye
-- gg sullivan
--
Lorenzo Cavallaro `Gigi Sullivan' <[EMAIL PROTECTED]> -- ITALY
Until I loved, life had no beauty;
I did not know I lived until I had loved. (Theodor Korner)
-
To unsubscribe from this list: send the line "unsubscribe linux-net" in
the body of a message to [EMAIL PROTECTED]
