LOGICAL OR WITH DOUBLE UNDERBAR·U+2A63

Character Information

Code Point
U+2A63
HEX
2A63
Unicode Plane
Basic Multilingual Plane
Category
Math Symbol

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 A9 A3
11100010 10101001 10100011
UTF16 (big Endian)
2A 63
00101010 01100011
UTF16 (little Endian)
63 2A
01100011 00101010
UTF32 (big Endian)
00 00 2A 63
00000000 00000000 00101010 01100011
UTF32 (little Endian)
63 2A 00 00
01100011 00101010 00000000 00000000
HTML Entity
⩣
URI Encoded
%E2%A9%A3

Description

The Unicode character U+2A63, known as LOGICAL OR WITH DOUBLE UNDERBAR, is a mathematical symbol that holds significant importance in digital text processing. Typically used in computer programming and logic gates, this character represents the logical OR operation between two binary values. It is an essential component in the realm of binary mathematics, where it serves to evaluate and manipulate data efficiently. While LOGICAL OR WITH DOUBLE UNDERBAR may not have a direct cultural or linguistic context, its technical usage underscores its critical role in various programming languages and algorithms. This symbol's accuracy and precision make it an indispensable tool for developers working with binary systems and digital logic.

How to type the symbol on Windows

Hold Alt and type 10851 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+2A63. 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+2A63 to binary: 00101010 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:
    11100010 10101001 10100011