LOGICAL AND WITH HORIZONTAL DASH·U+2A5C

Character Information

Code Point
U+2A5C
HEX
2A5C
Unicode Plane
Basic Multilingual Plane
Category
Math Symbol

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 A9 9C
11100010 10101001 10011100
UTF16 (big Endian)
2A 5C
00101010 01011100
UTF16 (little Endian)
5C 2A
01011100 00101010
UTF32 (big Endian)
00 00 2A 5C
00000000 00000000 00101010 01011100
UTF32 (little Endian)
5C 2A 00 00
01011100 00101010 00000000 00000000
HTML Entity
⩜
URI Encoded
%E2%A9%9C

Description

U+2A5C Logical AND with Horizontal Dash is a rare typographical character that serves an important purpose in certain digital text applications. Its primary role lies within the realm of formal logic systems, where it represents the logical conjunction "and" or truth-functional "AND." In these contexts, the character signifies that both operands must be true for the entire expression to hold true. While this Unicode character may not have a significant cultural or linguistic presence, it remains essential in technical applications where precise representation of logical operations is crucial. It helps to avoid potential misinterpretations and ambiguities in programming languages, digital text manipulation, and mathematical expressions that rely on accurate depiction of logical conjunctions.

How to type the symbol on Windows

Hold Alt and type 10844 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+2A5C. 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+2A5C to binary: 00101010 01011100. 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 10011100