How can I ignore part of an input in Java?

Question

I need to make a program that can convert hexadecimal values to decimal. I got the program to work but I need to make my program display the same output no matter if the hexadecimal value entered contains the "0x" in front or not.

this is my code.

import java.util.Scanner;

public class Main {

    public static long hex2decimal(String s){
        String digits = "0123456789ABCDEF";
        int val = 0;
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            int d = digits.indexOf(c);
            val = 16*val + d;
        }
        return val;
    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        long decimal;

        System.out.println("Please enter the hex string:");
        String s = sc.next().toUpperCase();
        decimal = hex2decimal(s);
        System.out.println(decimal);

    }
}

Possible duplicate of [Remove part of string](https://stackoverflow.com/questions/8694984/remove-part-of-string) — Logan, Jun 11 '18 at 02:50

score 1 · Answer 1 · answered Jun 11 '18 at 02:56

1

Why should you not using Integer.parseInt("121",16), rather custom logic for converting hex to decimal. Here 16 is radix telling number is hexadecimal and will be converted to decimal.

System.out.println(Integer.parseInt("121",16));

answered Jun 11 '18 at 02:56

Gaurav Srivastav

2,381
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`Integer.parseInt()` does not ignore `0x`, which I believe is what the asker is trying to do. – Logan Jun 11 '18 at 02:58
You can simply replace or ignore that before passing to parseInt – Gaurav Srivastav Jun 11 '18 at 03:00
Yes, but that is what the asker is trying to achieve — not the conversion part. I suggest adding it to your code. – Logan Jun 11 '18 at 03:02

How can I ignore part of an input in Java?

1 Answers1