What would be the best way (ideally, simplest) to convert an int to a binary string representation in Java?
For example, say the int is 156. The binary string representation of this would be "10011100".
19 Answers
Integer.toBinaryString(int i) There is also the java.lang.Integer.toString(int i, int base) method, which would be more appropriate if your code might one day handle bases other than 2 (binary). Keep in mind that this method only gives you an unsigned representation of the integer i, and if it is negative, it will tack on a negative sign at the front. It won't use two's complement.
0public static string intToBinary(int n) { String s = ""; while (n > 0) { s = ( (n % 2 ) == 0 ? "0" : "1") +s; n = n / 2; } return s; } One more way- By using java.lang.Integer you can get string representation of the first argument i in the radix (Octal - 8, Hex - 16, Binary - 2) specified by the second argument.
Integer.toString(i, radix) Example_
private void getStrtingRadix() { // TODO Auto-generated method stub /* returns the string representation of the unsigned integer in concern radix*/ System.out.println("Binary eqivalent of 100 = " + Integer.toString(100, 2)); System.out.println("Octal eqivalent of 100 = " + Integer.toString(100, 8)); System.out.println("Decimal eqivalent of 100 = " + Integer.toString(100, 10)); System.out.println("Hexadecimal eqivalent of 100 = " + Integer.toString(100, 16)); } OutPut_
Binary eqivalent of 100 = 1100100 Octal eqivalent of 100 = 144 Decimal eqivalent of 100 = 100 Hexadecimal eqivalent of 100 = 64 public class Main { public static String toBinary(int n, int l ) throws Exception { double pow = Math.pow(2, l); StringBuilder binary = new StringBuilder(); if ( pow < n ) { throw new Exception("The length must be big from number "); } int shift = l- 1; for (; shift >= 0 ; shift--) { int bit = (n >> shift) & 1; if (bit == 1) { binary.append("1"); } else { binary.append("0"); } } return binary.toString(); } public static void main(String[] args) throws Exception { System.out.println(" binary = " + toBinary(7, 4)); System.out.println(" binary = " + Integer.toString(7,2)); } } This is something I wrote a few minutes ago just messing around. Hope it helps!
public class Main { public static void main(String[] args) { ArrayList<Integer> powers = new ArrayList<Integer>(); ArrayList<Integer> binaryStore = new ArrayList<Integer>(); powers.add(128); powers.add(64); powers.add(32); powers.add(16); powers.add(8); powers.add(4); powers.add(2); powers.add(1); Scanner sc = new Scanner(System.in); System.out.println("Welcome to Paden9000 binary converter. Please enter an integer you wish to convert: "); int input = sc.nextInt(); int printableInput = input; for (int i : powers) { if (input < i) { binaryStore.add(0); } else { input = input - i; binaryStore.add(1); } } String newString= binaryStore.toString(); String finalOutput = newString.replace("[", "") .replace(" ", "") .replace("]", "") .replace(",", ""); System.out.println("Integer value: " + printableInput + "\nBinary value: " + finalOutput); sc.close(); } }
Convert Integer to Binary:
import java.util.Scanner; public class IntegerToBinary { public static void main(String[] args) { Scanner input = new Scanner( System.in ); System.out.println("Enter Integer: "); String integerString =input.nextLine(); System.out.println("Binary Number: "+Integer.toBinaryString(Integer.parseInt(integerString))); } } Output:
Enter Integer:
10
Binary Number: 1010
1Using built-in function:
String binaryNum = Integer.toBinaryString(int num); If you don't want to use the built-in function for converting int to binary then you can also do this:
import java.util.*; public class IntToBinary { public static void main(String[] args) { Scanner d = new Scanner(System.in); int n; n = d.nextInt(); StringBuilder sb = new StringBuilder(); while(n > 0){ int r = n%2; sb.append(r); n = n/2; } System.out.println(sb.reverse()); } } The simplest approach is to check whether or not the number is odd. If it is, by definition, its right-most binary number will be "1" (2^0). After we've determined this, we bit shift the number to the right and check the same value using recursion.
@Test public void shouldPrintBinary() { StringBuilder sb = new StringBuilder(); convert(1234, sb); } private void convert(int n, StringBuilder sb) { if (n > 0) { sb.append(n % 2); convert(n >> 1, sb); } else { System.out.println(sb.reverse().toString()); } } here is my methods, it is a little bit convince that number of bytes fixed
private void printByte(int value) { String currentBinary = Integer.toBinaryString(256 + value); System.out.println(currentBinary.substring(currentBinary.length() - 8)); } public int binaryToInteger(String binary) { char[] numbers = binary.toCharArray(); int result = 0; for(int i=numbers.length - 1; i>=0; i--) if(numbers[i]=='1') result += Math.pow(2, (numbers.length-i - 1)); return result; } Using bit shift is a little quicker...
public static String convertDecimalToBinary(int N) { StringBuilder binary = new StringBuilder(32); while (N > 0 ) { binary.append( N % 2 ); N >>= 1; } return binary.reverse().toString(); } This can be expressed in pseudocode as:
while(n > 0): remainder = n%2; n = n/2; Insert remainder to front of a list or push onto a stack Print list or stack You should really use Integer.toBinaryString() (as shown above), but if for some reason you want your own:
// Like Integer.toBinaryString, but always returns 32 chars public static String asBitString(int value) { final char[] buf = new char[32]; for (int i = 31; i >= 0; i--) { buf[31 - i] = ((1 << i) & value) == 0 ? '0' : '1'; } return new String(buf); } if the int value is 15, you can convert it to a binary as follows.
int x = 15; Integer.toBinaryString(x); if you have the binary value, you can convert it into int value as follows.
String binaryValue = "1010"; Integer.parseInt(binaryValue, 2); My 2cents:
public class Integer2Binary { public static void main(String[] args) { final int integer12 = 12; System.out.println(integer12 + " -> " + integer2Binary(integer12)); // 12 -> 00000000000000000000000000001100 } private static String integer2Binary(int n) { return new StringBuilder(Integer.toBinaryString(n)) .insert(0, "0".repeat(Integer.numberOfLeadingZeros(n))) .toString(); } } This should be quite simple with something like this :
public static String toBinary(int number){ StringBuilder sb = new StringBuilder(); if(number == 0) return "0"; while(number>=1){ sb.append(number%2); number = number / 2; } return sb.reverse().toString(); } public class BinaryConverter { public static String binaryConverter(int number) { String binary = ""; if (number == 1){ binary = "1"; return binary; } if (number == 0){ binary = "0"; return binary; } if (number > 1) { String i = Integer.toString(number % 2); binary = binary + i; binaryConverter(number/2); } return binary; } } In order to make it exactly 8 bit, I made a slight addition to @sandeep-saini 's answer:
public static String intToBinary(int number){ StringBuilder sb = new StringBuilder(); if(number == 0) return "0"; while(number>=1){ sb.append(number%2); number = number / 2; } while (sb.length() < 8){ sb.append("0"); } return sb.reverse().toString(); } So now for an input of 1 you get an output of 00000001
public static String intToBinaryString(int n) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < 32 && n != 0; i++) { sb.append((n&1) == 1 ? "1" : "0"); n >>= 1; } return sb.reverse().toString(); } We cannot use n%2 to check the first bit, because it's not right for negtive integer. We should use n&1.
