On Tue, Jul 29, 2014 at 1:59 AM, FRIGN <d...@frign.de> wrote: > Be careful: the n in ntohl does not stand for native, but network. > More or less by convention, the Network Byte Order today is BE. What > ntohl makes sure is that in case we are on a LE-system, it's arranged > properly. > ntohl is only superfluous here when we are on a BE-machine. [...]
That's what I was assuming. That on a BE machine, ntohl will be a no-op. Maybe I should spell out my reasoning here. The bytes in hdr are in Big Endian order. Unless I have my understanding of bit shifts backwards, this expression: (hdr[9] << 0) | (hdr[10] << 8) | (hdr[11] << 16) | (hdr[12] << 24) will result in a BE-ordered (network-ordered) value on LE machines, and a LE-ordered value on BE machines. Then, running it through ntohl: reverses the byte order on LE machines, which gives us an LE-ordered value (correct) or does nothing on BE machines, which gives us an LE-ordered value (wrong!). In any case, I feel like I'm making mountains out of molehills here, and this example code isn't used anywhere as far as I can see, so I'm not gonna bother you with this anymore.
