Unicode Han Character 'take by force, rob, snatch' (U+593A)

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U+593A browser display

take by force, rob, snatch

image of Unicode Han Character 'take by force, rob, snatch' (U+593A)
Raster image of U+593A

take by force, rob, snatch

Unicode Data

Name take by force, rob, snatch
Block CJK Unified Ideographs
Category Letter, Other [Lo]
Script Han (Hani)
Combine 0
BIDI Left-to-Right [L]
Version Unicode 1.1.0 (June, 1993)
Unicode Han Data
kIRG_GSource G0-3661
kPhonetic 277* 1388A*
kEACC 273940
kMainlandTelegraph 1161
kCangjie KDI
kHanyuPinlu duó(197)
kKangXi 0250.071
kTraditionalVariant U+596A
kRSUnicode 37.3
kXHC1983 0280.090:duó
kDefinition take by force, rob, snatch
kTGHZ2013 084.130:duó
kSMSZD2003Index 148.03 173.08
kUnihanCore2020 G
kCantonese dyut6
kIRGHanyuDaZidian 10527.010
kFourCornerCode 4034
kIICore AG
kGB0 2265
kTGH 2013:396
kHanYu 10527.010
kCCCII 273940
kMandarin duó
kTotalStrokes 6

Encodings

HTML Entity (decimal) 夺
HTML Entity (hex) 夺
How to type in Microsoft Windows Alt +593A
UTF-8 (hex) 0xE5 0xA4 0xBA (e5a4ba)
UTF-8 (binary) 11100101:10100100:10111010
UTF-16 (hex) 0x593A (593a)
UTF-16 (decimal) 22,842
UTF-32 (hex) 0x0000593A (593a)
UTF-32 (decimal) 22,842
C/C++/Java source code "\u593A"
Python source code u"\u593A"
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Java Data
string.toUpperCase() 夺
string.toLowerCase() 夺
Character.UnicodeBlock CJK_UNIFIED_IDEOGRAPHS
Character.charCount() 1
Character.getDirectionality() DIRECTIONALITY_LEFT_TO_RIGHT [0]
Character.getNumericValue() -1
Character.getType() 5
Character.isDefined() Yes
Character.isDigit() No
Character.isIdentifierIgnorable() No
Character.isISOControl() No
Character.isJavaIdentifierPart() Yes
Character.isJavaIdentifierStart() Yes
Character.isLetter() Yes
Character.isLetterOrDigit() Yes
Character.isLowerCase() No
Character.isMirrored() No
Character.isSpaceChar() No
Character.isSupplementaryCodePoint() No
Character.isTitleCase() No
Character.isUnicodeIdentifierPart() Yes
Character.isUnicodeIdentifierStart() Yes
Character.isUpperCase() No
Character.isValidCodePoint() Yes
Character.isWhitespace() No