no.
lets try example with p1=0x3ffff and p2=0x7ffff and assuming your m=0x7ffff mask
in tab2.
p1 (0x3ffff) = 0xf0 0xbf 0xbf 0xbf (0b11110000 0b10111111 0b10111111 0b10111111)
p2 (0x7ffff) = 0xf1 0xbf 0xbf 0xbf (0b11110001 0b10111111 0b10111111 0b10111111)
for m == 0x7ffff
(p1 & ~m) == 0
(p2 & ~m) == 0
if((p1 & ~m) != (p2 & ~m)) {
... bust
}
so this if case is never taken, tho the lead byte in the utf8 encoding is
different
for this p1-p2 range and the following encoded bytes are not comparable.
> 0xxxxxxx
> 110xxxxx 10mmmmmm
0x3f -> 6 bits (count m's)
> 1110xxxx 10mmmmmm 10mmmmmm
0xfff -> 12 bits (count m's)
> 11110xxx 10mmmmmm 10mmmmmm 10mmmmmm
0x3ffff -> 18 bits (count m's)
repeat example with 18 bit mask m=0x3ffff we get (p2 & ~m) == 0x40000 then
we bust and following encoded bytes are compared in each branch separately.
--
cinap
