Converting an int to a binary string representation in Java?
JavaStringBinaryIntJava Problem Overview
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".
Java Solutions
Solution 1 - Java
Integer.toBinaryString(int i)
Solution 2 - Java
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.
Solution 3 - Java
public static string intToBinary(int n)
{
String s = "";
while (n > 0)
{
s = ( (n % 2 ) == 0 ? "0" : "1") +s;
n = n / 2;
}
return s;
}
Solution 4 - Java
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
Solution 5 - Java
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));
}
}
Solution 6 - Java
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();
}
}
Solution 7 - Java
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
Solution 8 - Java
Using 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());
}
}
Solution 9 - Java
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());
}
}
Solution 10 - Java
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;
}
Solution 11 - Java
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();
}
Solution 12 - Java
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
Solution 13 - Java
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);
}
Solution 14 - Java
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);
Solution 15 - Java
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();
}
}
Solution 16 - Java
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();
}
Solution 17 - Java
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;
}
}
Solution 18 - Java
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
Solution 19 - Java
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.