TIFINAGH MODIFIER LETTER LABIALIZATION MARK·U+2D6F

Character Information

Code Point
U+2D6F
HEX
2D6F
Unicode Plane
Basic Multilingual Plane
Category
Modifier Letter

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 B5 AF
11100010 10110101 10101111
UTF16 (big Endian)
2D 6F
00101101 01101111
UTF16 (little Endian)
6F 2D
01101111 00101101
UTF32 (big Endian)
00 00 2D 6F
00000000 00000000 00101101 01101111
UTF32 (little Endian)
6F 2D 00 00
01101111 00101101 00000000 00000000
HTML Entity
ⵯ
URI Encoded
%E2%B5%AF

Description

U+2D6F TIFINAGH MODIFIER LETTER LABIALIZATION MARK is a specialized character in the Unicode standard, specifically designed to modify the pronunciation of certain letters within the Tifinagh script. This mark is used to denote labialization, meaning it indicates that a following letter should be pronounced with rounded lips. Typical usage of this character can be found in the representation and transcription of Berber languages spoken in North Africa, such as Tuareg, Tamazight, and Kabyle, where Tifinagh script is commonly employed. This mark holds a crucial role in preserving and promoting the linguistic heritage of these regions, allowing speakers to write their native languages with accurate phonetic representations. The Tifinagh Modifier Letter Labialization Mark contributes significantly to the cultural context of these communities by supporting the documentation, transmission, and learning of their unique languages.

How to type the symbol on Windows

Hold Alt and type 11631 on the numpad. Or use Character Map.

  1. Step 1: Determine the UTF-8 encoding bit layout

    The character has the Unicode code point U+2D6F. In UTF-8, it is encoded using 3 bytes because its codepoint is in the range of 0x0800 to 0xffff.

    Therefore we know that the UTF-8 encoding will be done over 16 bits within the final 24 bits and that it will have the format: 1110xxxx 10xxxxxx 10xxxxxx
    Where the x are the payload bits.

    UTF-8 Encoding bit layout by codepoint range
    Codepoint RangeBytesBit patternPayload length
    U+0000 - U+007F10xxxxxxx7 bits
    U+0080 - U+07FF2110xxxxx 10xxxxxx11 bits
    U+0800 - U+FFFF31110xxxx 10xxxxxx 10xxxxxx16 bits
    U+10000 - U+10FFFF411110xxx 10xxxxxx 10xxxxxx 10xxxxxx21 bits
  2. Step 2: Obtain the payload bits:

    Convert the hexadecimal code point U+2D6F to binary: 00101101 01101111. Those are the payload bits.

  3. Step 3: Fill in the bits to match the bit pattern:

    Obtain the final bytes by arranging the paylod bits to match the bit layout:
    11100010 10110101 10101111