BOPOMOFO LETTER ONG·U+31B2

Character Information

Code Point
U+31B2
HEX
31B2
Unicode Plane
Basic Multilingual Plane
Category
Other Letter

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E3 86 B2
11100011 10000110 10110010
UTF16 (big Endian)
31 B2
00110001 10110010
UTF16 (little Endian)
B2 31
10110010 00110001
UTF32 (big Endian)
00 00 31 B2
00000000 00000000 00110001 10110010
UTF32 (little Endian)
B2 31 00 00
10110010 00110001 00000000 00000000
HTML Entity
ㆲ
URI Encoded
%E3%86%B2

Description

The Unicode character U+31B2 represents the Bopomofo letter "Ong," which plays a crucial role in the phonetic transcription of Mandarin Chinese. In digital text, this character is primarily used to transcribe the sound represented by the Mandarin syllable "ong." The Bopomofo script, derived from the Latin alphabet, was created during the early 20th century and serves as a practical tool for teaching Mandarin pronunciation. This system is widely employed in Chinese-speaking regions to facilitate language learning and literacy development. As a result of its linguistic significance, U+31B2 holds substantial importance within typography and linguistics circles, highlighting the ongoing interplay between technological advancements and cultural preservation.

How to type the symbol on Windows

Hold Alt and type 12722 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+31B2. 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+31B2 to binary: 00110001 10110010. 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:
    11100011 10000110 10110010