From mboxrd@z Thu Jan  1 00:00:00 1970
Message-ID: <7cfbee84ff7ffe431e0974e4281cb0e5@felloff.net>
Date: Sun, 30 Mar 2014 20:05:27 +0200
From: cinap_lenrek@felloff.net
To: 9fans@9fans.net
In-Reply-To: <47ce4a1e83ccab956e28cc2b42fe2ba6@brasstown.quanstro.net>
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Subject: Re: [9fans] a strange bug in grep
Topicbox-Message-UUID: d2eda964-ead8-11e9-9d60-3106f5b1d025

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