LOGICAL AND WITH DOUBLE UNDERBAR·U+2A60

Character Information

Code Point
U+2A60
HEX
2A60
Unicode Plane
Basic Multilingual Plane
Category
Math Symbol

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 A9 A0
11100010 10101001 10100000
UTF16 (big Endian)
2A 60
00101010 01100000
UTF16 (little Endian)
60 2A
01100000 00101010
UTF32 (big Endian)
00 00 2A 60
00000000 00000000 00101010 01100000
UTF32 (little Endian)
60 2A 00 00
01100000 00101010 00000000 00000000
HTML Entity
⩠
URI Encoded
%E2%A9%A0

Description

The Unicode character U+2A60, known as the Logical AND with Double Underscore, is a specialized symbol used in digital text for various programming and mathematical applications. It is employed to denote a logical AND operation with additional emphasis due to the presence of two underscores (_) beneath it. This particular representation can be found in computer science and engineering contexts, where it serves as an operator that returns true only when both operands are true. Its role is crucial for binary decision-making processes in algorithms and programming languages. The character is not tied to any specific cultural or linguistic tradition, but rather serves a universal technical function within the Unicode Standard, which aims to represent all characters used across the world's written languages.

How to type the symbol on Windows

Hold Alt and type 10848 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+2A60. 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+2A60 to binary: 00101010 01100000. 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 10100000