Logic behind word-boundary difference between BSD grep and GNU grep

Can someone please explain/help me understand the logic behind this difference between BSD grep and GNU grep? I've added carets below the matches.

$ grep --version
grep (BSD grep) 2.5.1-FreeBSD
$ cat t1 
admin:*:80:root
$ grep '\<.' t1
admin:*:80:root
^^^^^   ^^ ^^^^

versus

$ grep --version
GNU grep 2.6.3

Copyright (C) 2009 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

$ cat t1 
admin:*:80:root
$ grep '\<.' t1
admin:*:80:root
^       ^  ^

It seems GNU grep is getting it right and BSD grep isn't. I thought \< is supposed to get the word boundary? Note grep '\<ad' t1 seems to get the same result in both versions, e.g.:

$ cat t2
admin:*:80:root
sadmin:*:81:root
$ grep '\<ad' t2
admin:*:80:root
^^

What gives?

Thanks in advance.

Richard

Solution

Using \> to specify word boundary wouldn't work for BSD grep. Try [[:<:]] instead.

This post gives pretty useful information on word boundaries for various utilities.