Converting some hexadecimal string to a decimal integer

Question

I wrote some code to convert my hexadecimal display string to decimal integer. However, when input is something like 100a or 625b (something with a letter) I got an error like this:

java.lang.NumberFormatException: For input string: " 100a" at java.lang.NumberFormatException.forInputString(Unknown Source) at java.lang.Integer.parseInt(Unknown Source)

How can I convert my string with letters to a decimal integer?

if(display.getText() != null)
{
    if(display.getText().contains("a") || display.getText().contains("b") ||
       display.getText().contains("c") || display.getText().contains("d") ||
       display.getText().contains("e") || display.getText().contains("f"))
    {
        temp1 = Integer.parseInt(display.getText(), 16);
        temp1 = (double) temp1;
    }
    else
    {
        temp1 = Double.parseDouble(String.valueOf(display.getText()));
    }
}

Answer 1

It looks like there's an extra (leading) space character in your string (" 100a"). You can use trim() to remove leading and trailing whitespaces:

temp1 = Integer.parseInt(display.getText().trim(), 16);

Or if you think the presence of a space means there's something else wrong, you'll have to look into it yourself, since we don't have the rest of your code.

Answer 2

public static int hex2decimal(String s) {
    String digits = "0123456789ABCDEF";
    s = s.toUpperCase();
    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;
}

That's the most efficient and elegant solution I have found on the Internet. Some of the other solutions provided here didn't always work for me.