Question about sha256 and AMD kernels

[Snoopy]

-a 3 -1 0123456789abcdef -2 0 ?1?1?1?1?1?1?1?1?2?2?2?2?2?2?2?2
or
-a 3 -1 0 ?h?h?h?h?h?h?h?h?1?1?1?1?1?1?1?1

should be basically the same, as "h" = 01234567890abcdef

sry but i still dont really get what you try to achieve

your given example (edit: the second example is the same)
-a 3 -1 0123456789abcdef -2 0 ?1?1?1?1?1?1?1?1?2?2?2?2?2?2?2?2
will result in passes like
afafafaf00000000
b1b2b3b400000000
(16 chars total, 8 zeros at the end, no spaces between)

but you said passes always end with  00 00 00 00 00 00 00 00 (these are 16 zeros, so your mask dont fit) the next thing, do these "spaces" belong to the password or not?

shortened example
sha256('ff 00 00') is something complete different than
sha256('ff0000')