# What does the ^ operator do in Java?

JavaOperatorsExponentiation## Java Problem Overview

What function does the `^`

(caret) operator serve in Java?

When I try this:

```
int a = 5^n;
```

...it gives me:
> for n = 5, returns 0

> for n = 4, returns 1

> for n = 6, returns 3

...so I guess it doesn't perform exponentiation. But what is it then?

## Java Solutions

## Solution 1 - Java

#### 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

#### 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 << k)`

is two to the *k*-th power for `k=0..63`

.

##### See also

> * 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
```

## Solution 2 - Java

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 & b and 3 & 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 << p left shift 3 << 2 12 Shifts the bits of n left p positions. Zero bits are shifted into the low-order positions.
n >> p right shift 5 >> 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 >>> p right shift -4 >>> 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.

## Solution 3 - Java

It's bitwise XOR, Java does not have an exponentiation operator, you would have to use `Math.pow()`

instead.

## Solution 4 - Java

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 => 1 0 1 (5)
1 0 0 (4)
----------
0 0 1 => 1
```

Similarly,

```
5 ^ 5 => 1 0 1 (5)
1 0 1 (5)
------------
0 0 0 => (0)
5 ^ 6 => 1 0 1 (5)
1 1 0 (6)
-----------
0 1 1 => 3
```

## Solution 5 - Java

It is the `XOR`

bitwise operator.

## Solution 6 - Java

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.

## Solution 7 - Java

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 ==> 0011 1100 (a)
0000 1101 (b)
------------- XOR
0011 0001 => 49
(a ^ b) will give 49 which is 0011 0001
```

## Solution 8 - Java

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.

## Solution 9 - Java

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) .

## Solution 10 - Java

To perform exponentiation, you can use Math.pow instead:

<https://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Math.html#pow%28double,%20double%29>

## Solution 11 - Java

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) .

## Solution 12 - Java

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

## Solution 13 - Java

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.

## Solution 14 - Java

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).

## Solution 15 - Java

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

## Solution 16 - Java

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.

## Solution 17 - Java

As an addition to the other answers, it's worth mentioning that the caret operator can also be used with boolean operands, and it returns true (if and only if) the operands are different:

```
System.out.println(true ^ true); // false
System.out.println(true ^ false); // true
System.out.println(false ^ false); // false
System.out.println(false ^ true); // true
```

## Solution 18 - Java

In other languages like Python you can do 10**2=100, try it.

## Solution 19 - Java

^ = (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