What function does the ^
(caret) operator serve in Java?
When I try this:
int a = 5^n;
...it gives me:
^
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!".
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
.
Merge note: this answer was merged from another question where the intention was to use exponentiation to convert a string
"8675309"
toint
without usingInteger.parseInt
as a programming exercise (^
denotes exponentiation from now on). The OP's intention was to compute8*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.
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