COPTIC OLD NUBIAN INDIRECT QUESTION MARK·U+2CFB

Character Information

Code Point
U+2CFB
HEX
2CFB
Unicode Plane
Basic Multilingual Plane
Category
Other Punctuation

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 B3 BB
11100010 10110011 10111011
UTF16 (big Endian)
2C FB
00101100 11111011
UTF16 (little Endian)
FB 2C
11111011 00101100
UTF32 (big Endian)
00 00 2C FB
00000000 00000000 00101100 11111011
UTF32 (little Endian)
FB 2C 00 00
11111011 00101100 00000000 00000000
HTML Entity
⳻
URI Encoded
%E2%B3%BB

Description

The character U+2CFB, known as the COPTIC OLD NUBIAN INDIRECT QUESTION MARK, is an important symbol in digital typography, particularly in languages such as Coptic and Old Nubian. This Unicode character serves a crucial role in these languages by helping to differentiate between direct and indirect questions, adding depth and clarity to the written text. It is also used in various digital applications, including software for linguistic research and translation services, where it plays a vital part in accurately conveying the nuances of these unique languages. The character's presence in the Unicode system reflects its significance in both historical and contemporary contexts, as well as its importance for preserving and promoting cultural diversity in written communication.

How to type the symbol on Windows

Hold Alt and type 11515 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+2CFB. 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+2CFB to binary: 00101100 11111011. 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 10110011 10111011