On Thu, Mar 8, 2018 at 1:42 PM, Jerry Gaffke via Groups.Io <jgaffke@...> wrote: Here's C code for machine A to send a 32 bit integer as a sequence of four bytes in little endian order:: It seems to me like Jerry is trying to point out that the left- and right-shift operators are endian-agnostic. They work in terms of the mathematically more-significant and less-significant directions without concern for the byte-order that an architecture uses to store its integers. so 256>>8 never gives a result of 0 on any architecture, even if the internal representation is 0x00 0x01. -Neil N0FN |