LEFTWARDS ARROW WITH DOTTED STEM·U+2B38

Character Information

Code Point
U+2B38
HEX
2B38
Unicode Plane
Basic Multilingual Plane
Category
Math Symbol

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 AC B8
11100010 10101100 10111000
UTF16 (big Endian)
2B 38
00101011 00111000
UTF16 (little Endian)
38 2B
00111000 00101011
UTF32 (big Endian)
00 00 2B 38
00000000 00000000 00101011 00111000
UTF32 (little Endian)
38 2B 00 00
00111000 00101011 00000000 00000000
HTML Entity
⬸
URI Encoded
%E2%AC%B8

Description

The Unicode character U+2B38, known as the "LEFTWARDS ARROW WITH DOTTED STEM", serves a specific function within digital text. It is often used to denote a reversal or direction change in mathematical equations, programming syntax, and logical expressions. This particular symbol carries no cultural or linguistic significance but holds great importance in technical contexts, such as computer science and engineering domains. Its unique appearance, featuring a leftward-facing arrow with a dotted stem, distinguishes it from other arrows commonly used in text, allowing for precise and unambiguous communication of directional changes or reversals within equations and algorithms.

How to type the symbol on Windows

Hold Alt and type 11064 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+2B38. 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+2B38 to binary: 00101011 00111000. 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 10101100 10111000