Reputation: 219
I want to convert my binary(which is in string) to hexadecimal string also, this is just a program fragment since this program is just a part of another bigger program:
//the variable name of the binary string is: "binary"
int digitNumber = 1;
int sum = 0;
int test = binary.length()%4;
if(test!=0) {
binary = padLeft(binary, test);
}
for(int i = 0; i < binary.length(); i++){
if(digitNumber == 1)
sum+=Integer.parseInt(binary.charAt(i) + "")*8;
else if(digitNumber == 2)
sum+=Integer.parseInt(binary.charAt(i) + "")*4;
else if(digitNumber == 3)
sum+=Integer.parseInt(binary.charAt(i) + "")*2;
else if(digitNumber == 4 || i < binary.length()+1){
sum+=Integer.parseInt(binary.charAt(i) + "")*1;
digitNumber = 0;
if(sum < 10)
System.out.print(sum);
else if(sum == 10)
System.out.print("A");
else if(sum == 11)
System.out.print("B");
else if(sum == 12)
System.out.print("C");
else if(sum == 13)
System.out.print("D");
else if(sum == 14)
System.out.print("E");
else if(sum == 15)
System.out.print("F");
sum=0;
}
digitNumber++;
}
public static String padLeft(String s, int n) {
return String.format("%0$"+n+"s", s);
}//i added this for padding
the problem is that i dont know if the padding works but i am sure that this program return a wrong hexadecimal conversion of the binary string I am trying to do this:
http://www.wikihow.com/Convert-Binary-to-Hexadecimal
PS: I need to implement it(not using any built-in function)
Upvotes: 17
Views: 81954
Reputation: 1
private final String[] hexValues = {"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"};
public void binaryToHexadecimal(String binary){
String hexadecimal;
binary = leftPad(binary);
System.out.println(convertBinaryToHexadecimal(binary));
}
public String convertBinaryToHexadecimal(String binary){
String hexadecimal = "";
int sum = 0;
int exp = 0;
for (int i=0; i<binary.length(); i++){
exp = 3 - i%4;
if((i%4)==3){
sum = sum + Integer.parseInt(binary.charAt(i)+"")*(int)(Math.pow(2,exp));
hexadecimal = hexadecimal + hexValues[sum];
sum = 0;
}
else
{
sum = sum + Integer.parseInt(binary.charAt(i)+"")*(int)(Math.pow(2,exp));
}
}
return hexadecimal;
}
public String leftPad(String binary){
int paddingCount = 0;
if ((binary.length()%4)>0)
paddingCount = 4-binary.length()%4;
while(paddingCount>0) {
binary = "0" + binary;
paddingCount--;
}
return binary;
}
Upvotes: 0
Reputation: 7722
Use this for any binary string length:
String hexString = new BigInteger(binaryString, 2).toString(16);
Upvotes: 11
Reputation: 1
/* * To change this license header, choose License Headers in Project Properties. * To change this template file, choose Tools | Templates * and open the template in the editor. */ package stringprocessing;
/** * * @author Zayeed Chowdhury */ public class StringProcessing {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
int index = 0;
String bin = "0000000101100101011011100110011100000001000000000000000010101010010101100110010101100011011010010110110101100001001000000100111001100101011101000111011101101111011100100110101101110011001000000100100001000001010100110010000001001001010100110101001101010101010001010100010000100000010000010010000001010010010001010101000101010101010010010101001001000101010001000010000001010111010001010100010101001011010011000101100100100000010101000100010101010011010101000010000001000110010011110101001000100000010101000100100001000101001000000100011001001111010011000100110001001111010101110100100101001110010001110010000001000011010011110101010101001110010101000100100101000101010100110010111101000001010100100100010101000001010100110011101000100000010100000110100101101110011000010110110000101100001000000100000101011010001110110010000001000001010101000010000000000001111000000011000100110010001110100011000100110011001000000101000001001101001000000100111101001110";
String[] hexString = new String[bin.length() / 4];
for (int i = 0; i < bin.length() / 4; i++) {
hexString[i] = "";
for (int j = index; j < index + 4; j++) {
hexString[i] += bin.charAt(j);
}
index += 4;
}
for (int i = 0; i < bin.length() / 4; i++) {
System.out.print(hexString[i] + " ");
}
System.out.println("\n" + bin.length());
String[] result = binaryToHex(hexString);
for (int i = 0; i < result.length; i++) {
System.out.print("" + result[i].toUpperCase());
}
System.out.println("");
}
public static String[] binaryToHex(String[] bin) {
String[] result = new String[bin.length];
for (int i = 0; i < bin.length; i++) {
result[i] = Integer.toHexString(Integer.parseInt(bin[i], 2));
}
//return Integer.toHexString(Integer.parseInt(bin[0], 2));
return result;
}
}
Upvotes: 0
Reputation:
By given binary number 01011011
, we will convert it at first to decimal number, each number will be Math.pow()
by decrementd length:
01011011 =(0 × 2(7)) + (1 × 2(6)) + (0 × 2(5)) + (1 × 2(4)) + (1 × 2(3)) + (0 × 2(2)) + (1 × 2(1)) + (1 × 2(0))
= (0 × 128) + (1 × 64) + (0 × 32) + (1 × 16) + (1 × 8) + (0 × 4) + (1 × 2) + (1 × 1)
= 0 + 64 + 0 + 16 + 8 + 0 + 2 + 1
= 91 (decimal form of binary number)
Now after get decimal number we have to convert it to hexa-decimal-number.
So, 91 is greater than 16. So, we have to divide by 16.
After dividing by 16, quotient is 5 and remainder is 11.
Remainder is less than 16.
Hexadecimal number of remainder is B.
Quotient is 5 and hexadecimal number of remainder is B.
That is, 91 = 16 × 5 +11 = B
5 = 16 × 0 + 5 = 5
=5B
Implementation:
String hexValue = binaryToHex(binaryValue);
//Display result
System.out.println(hexValue);
private static String binaryToHex(String binary) {
int decimalValue = 0;
int length = binary.length() - 1;
for (int i = 0; i < binary.length(); i++) {
decimalValue += Integer.parseInt(binary.charAt(i) + "") * Math.pow(2, length);
length--;
}
return decimalToHex(decimalValue);
}
private static String decimalToHex(int decimal){
String hex = "";
while (decimal != 0){
int hexValue = decimal % 16;
hex = toHexChar(hexValue) + hex;
decimal = decimal / 16;
}
return hex;
}
private static char toHexChar(int hexValue) {
if (hexValue <= 9 && hexValue >= 0)
return (char)(hexValue + '0');
else
return (char)(hexValue - 10 + 'A');
}
Upvotes: 0
Reputation: 1
import java.util.*;
public class BinaryToHexadecimal
{
public static void main()
{
Scanner sc=new Scanner(System.in);
System.out.println("enter the binary number");
double s=sc.nextDouble();
int c=0;
long s1=0;
String z="";
while(s>0)
{
s1=s1+(long)(Math.pow(2,c)*(long)(s%10));
s=(long)s/10;
c++;
}
while(s1>0)
{
long j=s1%16;
if(j==10)
{
z="A"+z;
}
else if(j==11)
{
z="B"+z;
}
else if(j==12)
{
z="C"+z;
}
else if(j==13)
{
z="D"+z;
}
else if(j==14)
{
z="E"+z;
}
else if(j==15)
{
z="F"+z;
}
else
{
z=j+z;
}
s1=s1/16;
}
System.out.println("The respective Hexadecimal number is : "+z);
}
}
Upvotes: 0
Reputation: 113
You can try something like this.
private void bitsToHexConversion(String bitStream){
int byteLength = 4;
int bitStartPos = 0, bitPos = 0;
String hexString = "";
int sum = 0;
// pad '0' to make input bit stream multiple of 4
if(bitStream.length()%4 !=0){
int tempCnt = 0;
int tempBit = bitStream.length() % 4;
while(tempCnt < (byteLength - tempBit)){
bitStream = "0" + bitStream;
tempCnt++;
}
}
// Group 4 bits, and find Hex equivalent
while(bitStartPos < bitStream.length()){
while(bitPos < byteLength){
sum = (int) (sum + Integer.parseInt("" + bitStream.charAt(bitStream.length()- bitStartPos -1)) * Math.pow(2, bitPos)) ;
bitPos++;
bitStartPos++;
}
if(sum < 10)
hexString = Integer.toString(sum) + hexString;
else
hexString = (char) (sum + 55) + hexString;
bitPos = 0;
sum = 0;
}
System.out.println("Hex String > "+ hexString);
}
Hope this helps :D
Upvotes: 3
Reputation: 393831
If you don't have to implement that conversion yourself, you can use existing code :
int decimal = Integer.parseInt(binaryStr,2);
String hexStr = Integer.toString(decimal,16);
If you must implement it yourself, there are several problems in your code :
sum
must be reset to 0 after each hex digit you output.System.out.print(digitNumber);
- here you should print sum
, not digitNumber
.Here's how the mostly fixed code looks :
int digitNumber = 1;
int sum = 0;
String binary = "011110101010";
for(int i = 0; i < binary.length(); i++){
if(digitNumber == 1)
sum+=Integer.parseInt(binary.charAt(i) + "")*8;
else if(digitNumber == 2)
sum+=Integer.parseInt(binary.charAt(i) + "")*4;
else if(digitNumber == 3)
sum+=Integer.parseInt(binary.charAt(i) + "")*2;
else if(digitNumber == 4 || i < binary.length()+1){
sum+=Integer.parseInt(binary.charAt(i) + "")*1;
digitNumber = 0;
if(sum < 10)
System.out.print(sum);
else if(sum == 10)
System.out.print("A");
else if(sum == 11)
System.out.print("B");
else if(sum == 12)
System.out.print("C");
else if(sum == 13)
System.out.print("D");
else if(sum == 14)
System.out.print("E");
else if(sum == 15)
System.out.print("F");
sum=0;
}
digitNumber++;
}
Output :
7AA
This will work only if the number of binary digits is divisable by 4, so you must add left 0
padding as a preliminray step.
Upvotes: 44