TAI LE LETTER A·U+1963

Character Information

Code Point
U+1963
HEX
1963
Unicode Plane
Basic Multilingual Plane
Category
Other Letter

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E1 A5 A3
11100001 10100101 10100011
UTF16 (big Endian)
19 63
00011001 01100011
UTF16 (little Endian)
63 19
01100011 00011001
UTF32 (big Endian)
00 00 19 63
00000000 00000000 00011001 01100011
UTF32 (little Endian)
63 19 00 00
01100011 00011001 00000000 00000000
HTML Entity
ᥣ
URI Encoded
%E1%A5%A3

Description

U+1963 is the Unicode code point for TAI LE LETTER A, a character from the Tai Le script used primarily in the Chamic languages such as Vietnamese and the Cham language of Vietnam. In digital text, it serves as an essential component in representing and encoding these linguistic systems accurately. The TAI LE LETTER A holds significance in both cultural and linguistic contexts, as it contributes to the preservation and promotion of the unique heritage of the Chamic languages. Its precise representation through Unicode also supports technical applications such as text processing, translation, and localization efforts for software and digital platforms. By ensuring accurate depiction of this character and others in its script, Unicode fosters a more inclusive and diverse global communication landscape.

How to type the symbol on Windows

Hold Alt and type 6499 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+1963. 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+1963 to binary: 00011001 01100011. 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:
    11100001 10100101 10100011