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: 88Explanation:
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 + 8 = 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.
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 numberstring 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 Codeint 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 numbersimport java.util.*; import java.util.HashMap; import java.util.Map; class GFG{ // Function to convert decimal no.// to hexadecimal numberstatic 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 int n = num; // use the same remainder technique. while (n != 0) { res = m.get(n % 16) + res; n /= 16; } } return res;} // Driver Codepublic 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 |
Hexa representation for 134 is 86 -1 is ffffffff -234 is ffffff16
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