Print an integer in binary format in Java

Question

I have a number and I want to print it in binary. I don't want to do it by writing an algorithm, Is there any built-in function for that in Java?

Answer 1

Here no need to depend only on binary or any other format... one flexible built in function is available That prints whichever format you want in your program.. Integer.toString(int,representation);

Integer.toString(100,8) // prints 144 --octal representation

Integer.toString(100,2) // prints 1100100 --binary representation

Integer.toString(100,16) //prints 64 --Hex representation

Answer 2

This is the simplest way of printing the internal binary representation of an integer. For Example: If we take n as 17 then the output will be: 0000 0000 0000 0000 0000 0000 0001 0001

void bitPattern(int n) {

        int mask = 1 << 31;
        int count = 0;
        while(mask != 0) {
            if(count%4 == 0)
                System.out.print(" ");

            if((mask&n) == 0) 

                System.out.print("0");



            else 
                System.out.print("1");


            count++;
            mask = mask >>> 1;


    }
    System.out.println();
}

Answer 3

I needed something to print things out nicely and separate the bits every n-bit. In other words display the leading zeros and show something like this:

n = 5463
output = 0000 0000 0000 0000 0001 0101 0101 0111

So here's what I wrote:

/**
 * Converts an integer to a 32-bit binary string
 * @param number
 *      The number to convert
 * @param groupSize
 *      The number of bits in a group
 * @return
 *      The 32-bit long bit string
 */
public static String intToString(int number, int groupSize) {
    StringBuilder result = new StringBuilder();

    for(int i = 31; i >= 0 ; i--) {
        int mask = 1 << i;
        result.append((number & mask) != 0 ? "1" : "0");

        if (i % groupSize == 0)
            result.append(" ");
    }
    result.replace(result.length() - 1, result.length(), "");

    return result.toString();
}

Invoke it like this:

public static void main(String[] args) {
    System.out.println(intToString(5463, 4));
}

Answer 4

    for(int i = 1; i <= 256; i++)
    {
        System.out.print(i + " "); //show integer
        System.out.println(Integer.toBinaryString(i) + " "); //show binary
        System.out.print(Integer.toOctalString(i) + " "); //show octal
        System.out.print(Integer.toHexString(i) + " "); //show hex

    }

Answer 5

Try this way:

public class Bin {
  public static void main(String[] args) {
    System.out.println(toBinary(0x94, 8));
  }

  public static String toBinary(int a, int bits) {
    if (--bits > 0)
        return toBinary(a>>1, bits)+((a&0x1)==0?"0":"1");
    else 
        return (a&0x1)==0?"0":"1";
  }

}

10010100

Answer 6

public static void main(String[] args) 
{
    int i = 13;
    short s = 13;
    byte b = 13;

    System.out.println("i: " + String.format("%32s", 
            Integer.toBinaryString(i)).replaceAll(" ", "0"));
    System.out.println("s: " + String.format("%16s", 
            Integer.toBinaryString(0xFFFF & s)).replaceAll(" ", "0"));
    System.out.println("b: " + String.format("%8s", 
            Integer.toBinaryString(0xFFFFFF & b)).replaceAll(" ", "0"));

}

Output:

i: 00000000000000000000000000001101
s: 0000000000001101
b: 00001101