Bitwise operations in Java: Test if in "1010101111011" a bit is set?

Question

Let's say, I have some user input (as a String) like "11010011011".

Now I want to check if a bit at a bit at a particular position is set (each digit should act as a flag).

Note: I am receiving the user's input as a String.

How can I do that?

Cant u use charAt(position) method ? And compare the value is 0/1? — Mukul Goel
– Mukul Goel, Commented Nov 1, 2012 at 15:03
I have no idea what would be the best way to do that. Maybe I would not need bitwise operation. — mrd
– mrd, Commented Nov 1, 2012 at 15:10

assylias · Accepted Answer · 2012-11-01 15:02:27Z

4

You could work with the string as is - say you want to check the first bit on the left:

if (input.charAt(0) == '1') { //

Alternatively if you want to work with a BitSet you can initialise it in a loop:

String input = "11010011011";
BitSet bs = new BitSet(input.length());
int i = 0;
for (char c : input.toCharArray()) {
    if (c == '1') bs.set(i);
    i++;
}

Then to check if the i-th bit is set:

boolean isSet = bs.get(i);

answered Nov 1, 2012 at 15:02

assylias

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Comments

iabdalkader · Accepted Answer · 2012-11-01 15:11:37Z

4

If you want to use bitwise operations, then first convert the string to integer and test with bitmasks:

int val = Integer.parseInt("11010011011", 2);
System.out.println(val & (1<<0)); //First bit
System.out.println(val & (1<<1)); //Second bit
System.out.println(val & (1<<2)); //Third bit
.....

edited Nov 1, 2012 at 15:11

answered Nov 1, 2012 at 15:06

iabdalkader

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