Generate bits out of int?

[Amandeep Singh]

int i = 11;

represented in bits as-

1 0 1 1

But how do we generate bits out of int in java?

[Bear Bibeault]

What's your algorithm? As the bits are the binary representation of the value, how would you generate the base 2 equivalent? In the US, this is 6th-grade math.

[Amandeep Singh]

Thanks for reply.

Following is my algorithm-

(1 * 2^3) + (0 * 2^2) + (1 * 2^1) + (1 * 2^0) = 8 + 0 + 2 + 1 = 11 

Yes, it is basics and I believe I learnt in 5th grade . I know on a piece of paper, I can calculate.

1 0 1 1

But in java, I'm looking for a method which changes it into machine readable bits. The algorithm which is used behind the scenes to manipulate the code or understand the int 11. I meant to say, I don't want to define my algorithm to calculate the bits. There should be inbuilt in java too to calculate the bits. What algorithm/method would be to calculate the bits?

Pardon me, if you think it is too stupid question.

[Claudiu Chelemen]

Doesn't the toBinaryString static method accomplish this ? eg:

int i = 11;
System.out.println(Integer.toBinaryString(i));

From there on, you can break it into byte[] or char[] easily.

Cheers!

[Hunter McMillen]

If you repeatedly divide the number by 2, the remainders will be the binary form of the number.

ex: # = 23

23 / 2 = 11 R 1 <- 1
11/2 = 5 R 1 <- 2
5 /2 = 2 R 1 <- 4
2/2 = 1 R 0
1/2 = 0 R 1 <- 16

10111 = 23


Hunter

Edit: reorder the numbers, I ordered them in reverse even though I wrote the correct values next to them.

[Rob Spoor]

Surely you mean 10111? Your algorithm works, you just need to add the new numbers to the start, not the end.