What does the ^ operator do in Java?

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执念已碎
执念已碎 2020-11-22 03:27

What function does the ^ (caret) operator serve in Java?

When I try this:

int a = 5^n;

...it gives me:

17条回答
  •  不思量自难忘°
    2020-11-22 03:48

    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
    

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