On 24 January 2014 07:16, Hans Petter Selasky <h...@bitfrost.no> wrote: > On 01/24/14 16:11, Adrian Chadd wrote: >> >> ... How's that matter? >> >> Adrian > > > Ok, read slowly: > > uint32_t x = 255U; > uint8_t y; > > On Big endian: > > memcpy(&y, &x, 1); > > y == 0; > > On Little endian: > > memcpy(&y, &x, 1); > > y == 255; > > If I'm not mistaken. The code is wrong because result depends on endianness > :-)
Right. But that has nothing to do with the memory copy operation. That has to do with how its stored. So again - how's memmove() not portable here? :) -a _______________________________________________ svn-src-head@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/svn-src-head To unsubscribe, send any mail to "svn-src-head-unsubscr...@freebsd.org"
