%e3%82%ab%e3%83%aa%e3%83%93%e3%82%a2%e3%83%b3%e3%82%b3%e3%83%a0 062212-055 !!top!! [ Linux ]

So taking E3 (0xEB) as first byte, first byte & 0x0F is 0x0B. Then second byte 82 & 0x3F is 0x02. Third byte ab & 0x3F is 0xAB. So code point is (0x0B << 12) | (0x02 << 6) | 0xAB = (0xB000) | 0x0200 | 0xAB = 0xB2AB.

Code point = (((first byte & 0x0F) << 12) | ((second byte & 0x3F) << 6) | (third byte & 0x3F)) So taking E3 (0xEB) as first byte, first byte & 0x0F is 0x0B

Starting with %E3%82%AB. Let me convert each of these sequences to ASCII. 12) | (0x02 &lt

Each %E3%82%AB is a three-byte sequence: So taking E3 (0xEB) as first byte, first byte & 0x0F is 0x0B