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?
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 << 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
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.
It's bitwise XOR, Java does not have an exponentiation operator, you would have to use Math.pow() instead.
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
It is the XOR bitwise operator.
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.
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
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.
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) .
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) .
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
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
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.
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).
^ is binary (as in base-2) xor, not exponentiation (which is not available as a Java operator). For exponentiation, see java.lang.Math.pow().
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.
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
^ = (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
In other languages like Python you can do 10**2=100, try it.
I see this comment, but don't understand it.
Get its last set bit
diff &= -diff;
I tried
int a = 3 & -3; it returns 1.
int a = 2 & -2; it returns 2.
int a = 4 & -4; it returns 4.
int a = 5 & -5; it returns 1.
The comment would be better expressed as 'Get the least significant bit set'. To understand what is going on, you need to examine how negative numbers are represented in binary. The technique is called twos complement and works by starting with the positive representation of the number; you complement each bit (i.e. 1 -> 0 and 0 -> 1). You then add 1 to this number. In the example of 12:
00001100 12
11110011 complement
00000001 binary 1
11110100 add to complement to form twos complement negative
If you now AND the original value with the negative, you get
00000100
where the only bit set corresponds to the least significant bit set in the original pattern.
As the comment said, diff & -diff returns the value of the last bit that was set on diff. For example:
diff = 14
.... = 1110 (binary)
.... ^ last set bit
.... 10 is the last set bit
.... 10 in decimal is 2
Another example
diff = 24
.... = 11000 (binary)
.... ^ last set bit
.... 1000 is the last set bit
.... 1000 in decimal is 8
I would recommend reading the guidelines for how to ask a well formulated question. One recommendation I can personally give is to have one sentence at the end of your question that sums up exactly what it is you want to know.
I executed this (~(1 << 3)) statement and I am getting -9 as a result.
Statement : (~(1 << 3))
Result : -9
Numbers in computers are stored in 2's complement form.
Your original number is 1,which is 0...0001 in binary. I'm skipping bit 4 to 30 as they all will be zero (consider 32-bit system).
Doing 1 << 3 will yield 0...1000 i.e. +8. In simple terms, it means multiply 1 by 2^i, where i = 3 here.
Now, inverting this yields 1111 1111 1111 1111 1111 1111 1111 1000 which is a negative number in 2's complement form.
To get the value of the number, drop the first 1(sign bit), invert the entire number again and add 1.
So, inverting again will give you 0...1000. Add 1 to this, so 0...1001.
This is the value of the number which is 9. And sign is negative because the first sign bit is 1.
<< is the left-shift operator, which in simple terms, when applied to a number, multiplies it by 2^i, where i is the number of bits to be shifted, for example :
1 << 3 = 8 (multiply 1 by 2^3)
2 << 4 = 32 (multiply 2 by 2^4)
and ~ is the NOT operator, which takes each bit in a number and toggles it. In simple terms, ~x = -x - 1 For example :
~100102 = 011012
~8 = -9
Now coming to your question, (~(1 << 3)) = (~8) = -9. For more info, check this answer : NOT(~) vs NEGATION(!).
I need a little explanation of the following statement, What is it doing?:
int result = 154 + (153 << 8) + (25 << 16) + (64 << 24);
/* what would be the value of result after this line and how it is calculated
Why we need this? */
(153 << 8) is equivalent to 153 * pow(2, 8)
You are actually shifting your bits towards left..
Also: -
(153 >> 8) is equivalent to 153 / pow(2, 8)
You can guess why.. This is actually shifting bits towards right..
E.G: -
3 => 0011
3 << 2 is equal to 1100 -> (Which is code for 12)
3 >> 2 is equal to 0000 -> (Which is code for 0) -> you can see - **(3 / 4 = 0)**
NOTE :- Note that the right shifting rounds off towards negative infinity..
For E.G: -
-3 >> 1 --> -2 (-1.5 is rounded to -2)
Lets see how it happens: -
In binary string representation: -
-3 ==> 11111111111111111111111111111101
-3 >> 1 ==> 11111111111111111111111111111110 (-3 / 2 = -1.5 -> rounded to -2)
(Note the left most bit is filled by the bit that was present at the left most end before shift (1 in this case))
So, the value becomes, -2 (for -3>>1, which is greater than -3)
This happens for negative numbers.. Right shifting a negative number gives a number greater than the original number..
Which is contrary to the positive number where the left most bit will be filled by 0... Thus value obtained will be less than the original..
3 ==> 00000000000000000000000000000011
3 >> 1 ==> 00000000000000000000000000000001 (3 / 2 = 1.5 -> rounded to 1)
(So left most bit remains 0. So, the value is 1 (less than 3), i.e., value is rounded off towards negative infinity and becomes 1 from 1.5)
Similarly you can devise results for left-shifting positive and negative number..
The answer is 1075419546. The left shift operator is basically addings 0s to the binary representations of the decimal numbers so this would simply to
154 + 153 * 2^8 + 25 * 2^16 + 64*2^24
So you can convert 153 to its binary representation, then add 8 zeros and convert back to decimal etc etc..
in practical terms, if you use the given operator on a math expression it has the same meaning as the following:
123 << n is exactly the same as 123 * 2 ^ n
For example, 2 << 2 is 2 * 2 ^ 2 which is 8 or the same as 1000
Otherwise, you are just shifting bits to the left:
3 = 11 then 3 << 2 = 1100.
Hope it makes it clear.
I have a variable that represents the XOR of 2 numbers. For example: int xor = 7 ^ 2;
I am looking into a code that according to comments finds the rightmost bit that is set in XOR:
int rightBitSet = xor & ~(xor - 1);
I can't follow how exactly does this piece of code work. I mean in the case of 7^2 it will indeed set rightBitSet to 0001 (in binary) i.e. 1. (indeed the rightmost bit set)
But if the xor is 7^3 then the rightBitSet is being set to 0100 i.e 4 which is also the same value as xor (and is not the rightmost bit set).
The logic of the code is to find a number that represents a different bit between the numbers that make up xor and although the comments indicate that it finds
the right most bit set, it seems to me that the code finds a bit pattern with 1 differing bit in any place.
Am I correct? I am not sure also how the code works. It seems that there is some relationship between a number X and the number X-1 in its binary representation?
What is this relationship?
The effect of subtracting 1 from a binary number is to replace the least significant 1 in it with a 0, and set all the less significant bits to 1. For example:
5 - 1 = 101 - 1 = 100 = 4
4 - 1 = 100 - 1 = 011 = 3
6 - 1 = 110 - 1 = 101 = 5
So in evaluating x & ~(x - 1): above x's least significant 1, ~(x - 1) has the same set bits as ~x, so above x's least significant 1, x & ~(x-1) has no 1 bits. By definition, x has a 1 bit at its least significant 1, and as we saw above ~(x - 1) will, too, but ~(x - 1) will have 0s below that point. Therefore, x & ~(x - 1) will have only one 1 bit, at the least significant bit of x.