Java Convert integer to hex integer

Question

I\'m trying to convert a number from an integer into an another integer which, if printed in hex, would look the same as the original integer.

For example:

C

Answer 1

int orig = 20;
int res = Integer.parseInt(""+orig, 16);

Answer 2

Simply do this:

public static int specialNum(num){

    return Integer.parseInt( Integer.toString(num) ,16)
}

It should convert any special decimal integer to its hexadecimal counterpart.

Answer 3

Another way to convert int to hex.

String hex = String.format("%X", int);

You can change capital X to x for lowercase.

Example:

String.format("%X", 31) results 1F.

String.format("%X", 32) results 20.

Answer 4

You could try something like this (the way you would do it on paper):

public static int solve(int x){
    int y=0;
    int i=0;

    while (x>0){
        y+=(x%10)*Math.pow(16,i);
        x/=10;
        i++;
    }
    return y;
}

public static void main(String args[]){
    System.out.println(solve(20));
    System.out.println(solve(54));
}

For the examples you have given this would calculate: 0*16^0+2*16^1=32 and 4*16^0+5*16^1=84

Answer 5

public static int convert(int n) {
  return Integer.valueOf(String.valueOf(n), 16);
}

public static void main(String[] args) {
  System.out.println(convert(20));  // 32
  System.out.println(convert(54));  // 84
}

That is, treat the original number as if it was in hexadecimal, and then convert to decimal.

Answer 6

The following is optimized iff you only want to print the hexa representation of a positive integer.

It should be blazing fast as it uses only bit manipulation, the utf-8 values of ASCII chars and recursion to avoid reversing a StringBuilder at the end.

public static void hexa(int num) {
    int m = 0;
    if( (m = num >>> 4) != 0 ) {
        hexa( m );
    }
    System.out.print((char)((m=num & 0x0F)+(m<10 ? 48 : 55)));
}