On x86 CPUs (well, 80386 and above), a word is 32 bits. A "word-aligned" piece of memory is one whose starting binary address ends in 00 (i.e. is a multiple of 4 bytes).
Also I thought the x86 unit of paging is 4K, not 8K, but maybe things have changed since the old 386 days. Cheers, -- Rod Roark, Sunset Systems http://www.sunsetsystems.com/ Offering preconfigured Linux computers, custom software and remote system administration services. Public Key: http://www.sunsetsystems.com/rodspublickey.asc On Tuesday 21 January 2003 11:20 am, Peter Jay Salzman wrote: > you'd think by now i'd know stuff like this. i'm embarrased to have to > ask this, but here it goes. > > i'm reading the man page for electric fence, and i'm not fully > understanding the sections on EF_ALIGNMENT and "WORD-ALIGNMENT AND > OVERRUN DETECTION". i feel like i "almost" understand them. > > i think i understand the concept of memory page as being the minimum > chunk of memory the kernel handles internally (8192 bytes minimum > allocation of memory on x86) and alignment, but i guess i don't know > what a word is. > > for example, the man page says that malloc() may be required to return > word aligned memory pages, so in the diagram: > > > 1 page allocated by malloc() > x ------------ > x+1 | | > x+2 | 8192 | > > | bytes on | > | x86 | > > ------------ > > i guess that places a restriction on what "x" is, but because i don't > know what a word is, i don't know what that restriction is. > > what's a word? :) > > or does it mean that there's a restriction on *size* of the page and not > the starting point? > > pete _______________________________________________ vox-tech mailing list [EMAIL PROTECTED] http://lists.lugod.org/mailman/listinfo/vox-tech
