What does the operator do in Java

Question

Can you please help me understand what the following code means   x    0 1

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It increases the value of the variable by the the value after     For example   float x   0  x    0 1    x is now 0 1 x    0 1    x is now 0 2   It s just a shorter version of   x   x 0 1

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It s bitwise XOR  Java does not have an exponentiation operator  you would have to use Math pow   instead

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devtop    Math pow x i  - mean  2   will add the result of the operation Math pow x i  - mean  2  to the devtop variable   A more simple example   int devtop   2  devtop    3     devtop now equals 5

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As many people have already pointed out  it s the XOR operator  Many people have also already pointed out that if you want exponentiation then you need to use Math pow   But I think it s also useful to note that   is just one of a family of operators that are collectively known as bitwise operators   Operator    Name         Example     Result  Description a  amp  b       and          3  amp  5       1       1 if both bits are 1  a   b       or           3   5       7       1 if either bit is 1  a   b       xor          3   5       6       1 if both bits are different   a          not           3          -4      Inverts the bits  n  lt  lt  p      left shift   3  lt  lt  2      12      Shifts the bits of n left p positions  Zero bits are shifted into the low-order positions  n  gt  gt  p      right shift  5  gt  gt  2      1       Shifts the bits of n right p positions  If n is a 2 s complement signed number  the sign bit is shifted into the high-order positions  n  gt  gt  gt  p     right shift  -4  gt  gt  gt  28   15      Shifts the bits of n right p positions  Zeros are shifted into the high-order positions    From here   These operators can come in handy when you need to read and write to integers where the individual bits should be interpreted as flags  or when a specific range of bits in an integer have a special meaning and you want to extract only those  You can do a lot of every day programming without ever needing to use these operators  but if you ever have to work with data at the bit level  a good knowledge of these operators is invaluable

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is binary  as in base-2  xor  not exponentiation  which is not available as a Java operator   For exponentiation  see java lang Math pow

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devtop    Math pow x i  - mean  2   adds Math pow x i  - mean  2  to devtop

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It is XOR operator  It is use to do bit operations on numbers  It has the behavior such that when you do a xor operation on same bits say 0 XOR 0   1 XOR 1 the result is 0  But if any of the bits is different then result is 1   So when you did 5 3 then you can look at these numbers 5  6 in their binary forms and thus the expression becomes  101  XOR  110  which gives the result  011  whose decimal representation is 3

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That is because you are using the xor operator   In java  or just about any other language    is bitwise xor  so of course   10   1   11  more info about bitwise operators  It s interesting how Java and C  don t have a power operator

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It is the XOR bitwise operator

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You can take a look at the bytecode whenever you want to understand how java operators work  Here if you compile   int x   0  x    0 1    the bytecode will be accessible with jdk command javap -c    class   you can refer to Java bytecode instruction listings for more explanation about bytecode   0  iconst 0     load the int value 0 onto the stack 1  istore 1     store int value into variable 1  x  2  iload 1    load an int value from local variable 1  x  3  i2d    convert an int into a double  cast x to double  4  ldc2 w         2                     double 0 1d - gt  push a constant value  0 1  from a constant pool onto the stack 7  dadd     add two doubles  pops two doubles from stack  adds them  and pushes the answer onto stack  8  d2i    convert a double to an int  pops a value from stack  casts it to int and pushes it onto stack  9  istore 1    store int value into variable 1  x    Now it is clear that java compiler promotes x to double and then adds it with 0 1  Finally it casts the answer to integer   There is one interesting fact I found out that when you write   byte b   10  b    0 1    compiler casts b to double  adds it with 0 1  casts the result which is double to integer  and finally casts it to byte and that is because there is no instruction to cast double to byte directly  You can check the bytecode if you doubt

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As already stated by the other answer s   it s the  exclusive or   XOR  operator  For more information on bit-operators in Java  see  http   java sun com docs books tutorial java nutsandbolts op3 html

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AraK s link points to the definition of exclusive-or  which explains how this function works for two boolean values   The missing piece of information is how this applies to two integers  or integer-type values   Bitwise exclusive-or is applied to pairs of corresponding binary digits in two numbers  and the results are re-assembled into an integer result   To use your example    The binary representation of 5 is 0101  The binary representation of 4 is 0100    A simple way to define bitwise XOR is to say the result has a 1 in every place where the two input numbers differ   With 4 and 5  the only difference is in the last place  so  0101   0100   0001  5   4   1

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The   operator in Java    in Java is the exclusive-or   xor   operator   Let s take 5 6 as example    decimal      binary       5        101      6        110 ------------------ xor      3        011   This the truth table for bitwise  JLS 15 22 1  and logical  JLS 15 22 2  xor       0 1          F T -- -----     -- ----- 0   0 1      F   F T 1   1 0      T   T F   More simply  you can also think of xor as  this or that  but not both     See also   Wikipedia  exclusive-or     Exponentiation in Java  As for integer exponentiation  unfortunately Java does not have such an operator  You can use double Math pow double  double   casting the result to int if necessary    You can also use the traditional bit-shifting trick to compute some powers of two  That is   1L  lt  lt  k  is two to the k-th power for k 0  63   See also   Wikipedia  Arithmetic shift        Merge note  this answer was merged from another question where the intention was to use exponentiation to convert a string  8675309  to int without using Integer parseInt as a programming exercise     denotes exponentiation from now on   The OP s intention was to compute 8 10 6   6 10 5   7 10 4   5 10 3   3 10 2   0 10 1   9 10 0   8675309  the next part of this answer addresses that exponentiation is not necessary for this task    Horner s scheme  Addressing your specific need  you actually don t need to compute various powers of 10  You can use what is called the Horner s scheme  which is not only simple but also efficient   Since you re doing this as a personal exercise  I won t give the Java code  but here s the main idea   8675309   8 10 6   6 10 5   7 10 4   5 10 3   3 10 2   0 10 1   9 10 0                8 10   6  10   7  10   5  10   3  10   0  10   9   It may look complicated at first  but it really isn t  You basically read the digits left to right  and you multiply your result so far by 10 before adding the next digit   In table form   step   result  digit  result 10 digit    1   init 0      8                8    2        8      6               86    3       86      7              867    4      867      5             8675    5     8675      3            86753    6    86753      0           867530    7   867530      9          8675309 final

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XOR operator rule     0   0   0 1   1   0 0   1   1 1   0   1   Binary representation of 4  5 and 6    4   1 0 0  5   1 0 1 6   1 1 0   now  perform XOR operation on 5 and 4         5   4   gt  1  0  1    5                1  0  0    4              ----------               0  0  1     gt  1   Similarly    5   5   gt  1   0   1     5           1   0   1     5         ------------          0   0   0     gt   0    5   6   gt  1   0   1   5           1   1   0   6          -----------          0   1   1    gt  3

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x    y is x   x   y x -  y is x   x - y x    y is x   x   y x    y is x   x   y x    y is x   x   y x    y is x   x   y x  amp   y is x   x  amp  y x    y is x   x   y   and so on

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It s one of the assignment operators  It takes the value of x  adds 0 1 to it  and then stores the result of  x   0 1  back into x   So   double x   1 3  x    0 1        sets  x  to 1 4   It s functionally identical to  but shorter than   double x   1 3  x   x   0 1    NOTE  When doing floating-point math  things don t always work the way you think they will

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You can use Math pow instead   https   docs oracle com javase 1 5 0 docs api java lang Math html pow 28double  20double 29

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The  common knowledge  of programming is that x    y is an equivalent shorthand notation of x   x   y  As long as x and y are of the same type  for example  both are ints   you may consider the two statements equivalent   However  in Java  x    y is not identical to x   x   y in general    If x and y are of different types  the behavior of the two statements differs due to the rules of the language  For example  let s have x    0  int  and y    1 1  double        int x   0      x    1 1        just fine  hidden cast  x    1 after assignment     x   x   1 1     won t compile   cannot convert from double to int       performs an implicit cast  whereas for   you need to explicitly cast the second operand  otherwise you d get a compiler error   Quote from Joshua Bloch s Java Puzzlers            compound assignment expressions automatically cast the result of   the computation they perform  to the type of the variable on their   left-hand side  If the type of the result is identical to the type of   the variable  the cast has no effect  If  however  the type of the   result is wider than that of the variable   the  compound    assignment  operator  performs  a  silent  narrowing  primitive   conversion  JLS 5 1 3

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It is the Bitwise xor operator in java which results 1 for different value of bit  ie 1   0   1  and 0 for same value of bit  ie 0   0   0  when a number is written in binary form   ex  -   To use your example   The binary representation of 5 is 0101  The binary representation of 4 is 0100   A simple way to define Bitwise XOR is to say the result has a 1 in every place where the two input numbers differ   0101   0100   0001  5   4   1

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XOR operator rule  0   0   0 1   1   0 0   1   1 1   0   1   Bitwise operator works on bits and performs bit-by-bit operation  Assume if a   60 and b   13  now in binary format they will be as follows -  a   0011 1100  b   0000 1101    a b    gt  0011 1100   a          0000 1101   b          -------------  XOR         0011 0001    gt  49   a   b  will give 49 which is 0011 0001

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Lot many people have already explained about what it is and how it can be used but apart from the obvious you can use this operator to do a lot of programming tricks like   XORing of all the elements in a boolean array would tell you if the array has odd number of true elements If you have an array with all numbers repeating even number of times except one which repeats odd number of times you can find that by XORing all elements  Swapping values without using temporary variable Finding missing number in the range 1 to n Basic validation of data sent over the network    Lot many such tricks can be done using bit wise operators  interesting topic to explore

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It is the bitwise xor operator in java which results 1 for different value  ie 1   0   1  and 0 for same value  ie 0   0   0

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As others have said  it s bitwise XOR  If you want to raise a number to a given power  use Math pow a   b   where a is a number and b is the power

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In other languages like Python you can do 10  2 100  try it

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In java the default type of numbers like 2 or -2 without a fractional component  is int and unlike c  that s not an object and we can t do sth like 2 tostring as in c  and the default type of numbers like 2 5 with a fractional component  is double  So if you write   short s   2  s   s   4    you will get a compilation error that int cannot be cast into short also if you do sth like below   float f   4 6  f   f   4 3    you will get two compilation errors for setting double  4 6  to a float variable at both lines and the error of first line is logical because float and double use different system of storing numbers and using one instead of another can cause data loss  two examples mentioned can be changed like this   s    4 f    4 3   which both have an implicit cast behind code and have no compile errors  Another point worthy of consideration is numbers in the range of  byte  data type are cached in java and thus numbers -128 to 127 are of type byte in java and so this code doesn t have any compile errors   byte b   127   but this one has an error indeed   byte b   128   because 128 is an int in java  about long numbers we are recommended to use an L after the number for the matter of integer overflow like this   long l   2134324235234235L   in java we don t have operator overloading like c   but    is overloaded only for String and not for the let s say StringBuilder or StringBuffer and we can use it instead of String  concat  method but as we know String is immutable and that will make another object and will not change the same object as before     String str    Hello   str     World     It s fine

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bitwise XOR  Description  Binary XOR Operator copies the bit if it is set in one operand but not both   example   A   B  will give 49 which is 0011 0001

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