十六进制到十进制转换问题

Question

first post here. I have an assignment to convert from hex to decimal, including other conversions but this one has me stumped...

So the hex values I need to convert are 12345678, 2A3DF8A7, 00FF00FF. 00FF00FF works fine, it's the number values that are giving me trouble.

I know you're supposed to multiply the digit by 16^n, but for some reason I'm getting values that are way too high even though I think I am doing it right. Obviously not since it's not working. Help would be greatly appreciated!

My code has changed throughout the last hour messing with it, but it's the last "else" now that is the problem.

public void toDec()
{
    dec = 0;
    for (int j = 0, i = 7; j < hex.length(); j++){

        if (hex.charAt(j) == 'A') {
            dec += (10 * (int)Math.pow(16, i));
        }
        else if (hex.charAt(j) == 'B') {
            dec += (11 * (int)Math.pow(16, i));
        }
        else if (hex.charAt(j) == 'C') {
            dec += (12 * (int)Math.pow(16, i));
        }
        else if (hex.charAt(j) == 'D') {
            dec += (13 * (int)Math.pow(16, i));
        }
        else if (hex.charAt(j) == 'E') {
            dec += (14 * (int)Math.pow(16, i));
        }
        else if (hex.charAt(j) == 'F') {
            dec += (15 * (int)Math.pow(16, i));
        }
        else if (hex.charAt(j) == '0') {
            dec = dec;
        }
        else {
            dec += ((int)hex.charAt(j)) * ((int)Math.pow(16, i));
        }
        i--;
    }
}

Answer 1

Your problem is in your else clause, handling the digits 1 to 9.

The problem is that you're confusing the digits with the characters representing them. For example, the digit 1 is represented by the character '1' , which has a character code of 49. So, when you cast a '1' character to int in your else clause (ie (int)hex.charAt(j) ), what you get back is it's character code of 49, not 1.

Now, the easy (and idiomatic) way to fix this is to use integer math on your characters. For example, use hex.charAt(j) - '0' instead of (int)hex.charAt(j) . This works because the characters '0' through '9' are assigned consecutive character codes 48 through 57, so (for example) subtracting the character code for '0' (which is 48) from the character code for '1' (which is 49) returns the actual digit value, since 49 - 48 = 1 . The same works for the other digits, too: '5' - '0' is equivalent to 53 - 48 , which is (of course) equal to 5.

Even better, the same approach also works for your other hexadecimal characters too! For example, if you do 'C' - 'A' + 10 , you'll get the result 12 (which is the hexadecimal value of the letter C).

Answer 2

Casting char to an int is the problem here - instead of int value, representing specific digit, you get char code. Just give it a try: System.out.println((int) '5');

The easiest way to solve that would be (hex.charAt(j) - '0') .