Arnob - 2 months ago 11x

Java Question

I have the following code that computes the parity of a binary word:

`public class Parity1 {`

public static short parity(long x) {

short result = 0;

while (x != 0) {

result ^= (x & 1);

x >>>= 1;

}

return result;

}

If I have odd numbers of ones like 1011, I will get 1 and otherwise like 100001000, I will get 0.However, I am not seeing how the code is working.If I take 1011, and I pass it then I get the following:

`number:1011`

result : 0

result xor (1&1) == 1=>result

x >>>=1 => 0101

----------------------

number:0101

result : 1

result xor (1&1) == 0 => result

x >>>=1 => 0010

----------------------

number :0010

result : 0

result xor (0&1) ==0 => result

x >>>=1 => 0001

------------------------

number :0001

result : 0

result xor (1&1) ==1 => result

x >>>=1 => 0000

------------------------

number :0000

result : 1

result xor (1&1) ==1 => result

while (x!=0) but x(number) is 0

ends with result 0.

Can someone please explain what I am missing?

Here is another method that does the same thing but efficiently. However,I'm not getting it too either.

`public static short parity(long x){`

short result =0;

while (x!=0){

result ^=1;

x &= (x-1);

}

return result;

}

Answer

The code that you've posted looks correct, which leads me to think that you might be invoking it incorrectly. You gave 1011 as an example of a number where the parity should be 1 even though you're getting 0. If you use the *binary* number 1011 (decimal: 11), then you should get 1, but if you use the *decimal* number 1,011 (one thousand eleven), then you *should* get 0 because its binary representation is

```
1111110011
```

which has an even number of 1s in it.

If you want to invoke the function with the binary number 1011, invoke it as

```
parity(0b1011)
```

rather than

```
parity(1011).
```

Source (Stackoverflow)

Comments