>
> Baruch Siach <[email protected]> wrote on 2011/09/08 07:06:27:
> >
> > Hi Joakim,
>
> ...
>
> > > if (nbytes & 1) /* Odd */
> > > sum += *(u_char *)ptr; /* one byte only */
> >
> > This breaks big endian systems like PowerPC. The assignment via pointer
> > dance
> > of the original implementation is required, because we need the last byte to
> > be considered as MSB of a 16bit word, as if we had one more '\0' byte in
> > buffer. Note that the generated checksum is the same on big and little
> > endian
> > machines in terms of memory representation. They differ however in their
> > numeric representation. So, for example the checksum of
> >
> > uint8_t buf[] = {0x12, 0x34, 0x56};
> >
> > is 0xcb97 on little endian machines, and 0x97cb on big endian machines. This
> > is OK since we use the memory representation of the network packet.
>
> Ouch, even RFC 1071 got it wrong then.
I just checked and all this ptr casting confuses gcc and generates
bad code. Using an union is better, both for ppc and x86.
This is only compile tested:
if (len & 1) {
union uu {
uint16_t val;
uint8_t b[2];
} tmp = {0};
/* Make sure that the left-over byte is added correctly both
* with little and big endian hosts */
tmp.b[0] = *(uint8_t*)w;
sum += tmp.val;
}
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