Skip to content
Related Articles
Open in App
Not now

Related Articles

Convert Decimal To Hexa-Decimal including negative numbers

Improve Article
Save Article
  • Last Updated : 30 May, 2022
Improve Article
Save Article

Given a number N in decimal format, the task is to convert it to the hexadecimal representation of N as a string. Negative numbers are stored in 2’s complement form.
Examples: 
 

Input: N = 134 
Output: 86
Explanation: 
134 = 00000000000000000000000010001000 in 32 bit representation. Grouping in four-size chunks and converting each chunk to equivalent hexadecimal yields 88. Also, we can see 8*16 + 6 = 134. We will also get the same result by remainder technique discussed in other post.
Input: N = -1 
Output: ffffffff 
 

Approach: 
The ides is to store negative numbers in a bigger size to trick the compiler to read it as positive instead of negative and then use the normal remainder technique. Store num in a u_int, size of u_it is greater, it will be positive since MSB is 0.

For Java there is no unsigned int data type. So in case of java, convert number to long and make the 32 higher bits as all zeroes. Idea remains the same as above.

num =  (long)Math.pow(2, 32) + num ;
Below is the implementation of the above approach: 
 

C++




// C++ program to convert decimal
// to hexadecimal covering negative numbers
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to convert decimal no.
// to hexadecimal number
string Hex(int num)
{
    // map for decimal to hexa, 0-9 are
    // straightforward, alphabets a-f used
    // for 10 to 15.
    map<int, char> m;
 
    char digit = '0';
    char c = 'a';
 
    for (int i = 0; i <= 15; i++) {
        if (i < 10) {
            m[i] = digit++;
        }
        else {
            m[i] = c++;
        }
    }
 
    // string to be returned
    string res = "";
 
    // check if num is 0 and directly return "0"
    if (!num) {
        return "0";
    }
    // if num>0, use normal technique as
    // discussed in other post
    if (num > 0) {
        while (num) {
            res = m[num % 16] + res;
            num /= 16;
        }
    }
    // if num<0, we need to use the elaborated
    // trick above, lets see this
    else {
        // store num in a u_int, size of u_it is greater,
        // it will be positive since msb is 0
        u_int n = num;
 
        // use the same remainder technique.
        while (n) {
            res = m[n % 16] + res;
            n /= 16;
        }
    }
 
    return res;
}
 
// Driver Code
int main()
{
    int x = 134, y = -1, z = -234;
 
    cout << "Hexa representation for" << endl;
    cout << x << " is " << Hex(x) << endl;
    cout << y << " is " << Hex(y) << endl;
    cout << z << " is " << Hex(z) << endl;
 
    return 0;
}

Java




// Java program to convert decimal
// to hexadecimal covering negative numbers
import java.util.*;
import java.util.HashMap;
import java.util.Map;
 
class GFG
{
 
// Function to convert decimal no.
// to hexadecimal number
static String Hex(int num)
{
    // map for decimal to hexa, 0-9 are
    // straightforward, alphabets a-f used
    // for 10 to 15.
     
    HashMap<Integer, Character> m = new HashMap<Integer, Character>();
 
    char digit = '0';
    char c = 'a';
 
    for (int i = 0; i <= 15; i++) {
        if (i < 10) {
            m.put(i, digit);
            digit++;
        }
        else {
            m.put(i, c);
            c++;
        }
    }
 
    // string to be returned
    String res = "";
 
    // check if num is 0 and directly return "0"
    if (num == 0) {
        return "0";
    }
    // if num>0, use normal technique as
    // discussed in other post
    if (num > 0) {
        while (num != 0) {
            res = m.get(num % 16) + res;
            num /= 16;
        }
    }
    // if num<0, we need to use the elaborated
    // trick above, lets see this
    else {
        // store num in a u_int, size of u_it is greater,
        // it will be positive since msb is 0
        long n = num;
        num =  (long)Math.pow(2, 32) + num ;
        // use the same remainder technique.
        while (n != 0) {
            res = m.get(n % 16) + res;
            n /= 16;
        }
    }
 
    return res;
}
 
// Driver Code
public static void main(String []args)
{
    int x = 134, y = -1, z = -234;
 
    System.out.println("Hexa representation for" );
    System.out.println(x +" is " + Hex(x));
    System.out.println( y +" is " + Hex(y));
    System.out.println( z + " is " + Hex(z));   
}
}
 
// This code is contributed by chitranayal

Python3




# Python3 program to convert decimal
# to hexadecimal covering negative numbers
 
# Function to convert decimal no.
# to hexadecimal number
def Hex(num) :
 
    # map for decimal to hexa, 0-9 are
    # straightforward, alphabets a-f used
    # for 10 to 15.
    m = dict.fromkeys(range(16), 0);
 
    digit = ord('0');
    c = ord('a');
 
    for i in range(16) :
        if (i < 10) :
            m[i] = chr(digit);
            digit += 1;
         
        else :
            m[i] = chr(c);
            c += 1
 
    # string to be returned
    res = "";
 
    # check if num is 0 and directly return "0"
    if (not num) :
        return "0";
 
    # if num>0, use normal technique as
    # discussed in other post
    if (num > 0) :
        while (num) :
            res = m[num % 16] + res;
            num //= 16;
     
    # if num<0, we need to use the elaborated
    # trick above, lets see this
    else :
         
        # store num in a u_int, size of u_it is greater,
        # it will be positive since msb is 0
        n = num + 2**32;
 
        # use the same remainder technique.
        while (n) :
            res = m[n % 16] + res;
            n //= 16;
 
    return res;
 
# Driver Code
if __name__ == "__main__" :
 
    x = 134; y = -1; z = -234;
 
    print("Hexa representation for");
    print(x, "is", Hex(x));
    print(y, "is", Hex(y));
    print(z, "is", Hex(z));
 
# This code is contributed by AnkitRai01

C#




// C# program to convert decimal
// to hexadecimal covering negative numbers
using System;
using System.Collections.Generic;
 
public class GFG {
 
  // Function to convert decimal no.
  // to hexadecimal number
  static string Hex(long num)
  {
 
    // map for decimal to hexa, 0-9 are
    // straightforward, alphabets a-f used
    // for 10 to 15.
    IDictionary<long, char> m
      = new Dictionary<long, char>();
 
    char digit = '0';
    char c = 'a';
 
    for (int i = 0; i <= 15; i++) {
      if (i < 10) {
        m[i] = digit;
        digit++;
      }
      else {
        m[i] = c;
        c++;
      }
    }
 
    // string to be returned
    string res = "";
 
    // check if num is 0 and directly return "0"
    if (num == 0) {
      return "0";
    }
    // if num>0, use normal technique as
    // discussed in other post
    if (num > 0) {
      while (num != 0) {
        res = m[num % 16] + res;
        num /= 16;
      }
    }
    // if num<0, we need to use the elaborated
    // trick above, lets see this
    else {
      // we shall convert num to a 32 bit number
      num = (long)Math.Pow(2, 32) + num;
      long n = num;
      // use the same remainder technique.
      while (n != 0) {
        res = m[n % 16] + res;
        n /= 16;
      }
    }
 
    return res;
  }
 
  public static void Main(string[] args)
  {
    long x = 134, y = -1, z = -234;
 
    Console.WriteLine("Hexa representation for");
    Console.WriteLine(x + " is " + Hex(x));
    Console.WriteLine(y + " is " + Hex(y));
    Console.WriteLine(z + " is " + Hex(z));
  }
}
 
// this code is contributed by phasing17

Javascript




<script>
 
// JavaScript program to convert decimal
// to hexadecimal covering negative numbers
 
// Function to convert decimal no.
// to hexadecimal number
function Hex(num)
{
    // map for decimal to hexa, 0-9 are
    // straightforward, alphabets a-f used
    // for 10 to 15.
       
    let m = new Map();
   
    let digit = '0'.charCodeAt(0);
    let c = 'a'.charCodeAt(0);
   
    for (let i = 0; i <= 15; i++) {
        if (i < 10) {
            m.set(i, String.fromCharCode(digit));
            digit++;
        }
        else {
            m.set(i, String.fromCharCode(c));
            c++;
        }
    }
   
    // string to be returned
    let res = "";
   
    // check if num is 0 and directly return "0"
    if (num == 0) {
        return "0";
    }
    // if num>0, use normal technique as
    // discussed in other post
    if (num > 0) {
        while (num != 0) {
            res = m.get(num % 16) + res;
            num =Math.floor(num/ 16);
        }
    }
    // if num<0, we need to use the elaborated
    // trick above, lets see this
    else {
        // store num in a u_int, size of u_it is greater,
        // it will be positive since msb is 0
        let n = num+Math.pow(2,32);
   
        // use the same remainder technique.
        while (n != 0) {
            res = m.get(n % 16) + res;
            n = Math.floor(n/16);
        }
    }
   
    return res;
}
 
// Driver Code
let x = 134, y = -1, z = -234;
document.write("Hexa representation for<br>" );
document.write(x +" is " + Hex(x)+"<br>");
document.write( y +" is " + Hex(y)+"<br>");
document.write( z + " is " + Hex(z)+"<br>");
 
 
// This code is contributed by rag2127
 
</script>

Output: 

Hexa representation for
134 is 86
-1 is ffffffff
-234 is ffffff16

 

Time Complexity: O(log16num)

Auxiliary Space: O(15)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!