Converting string to int

Question

I cannot convert this string to integer.
int tempI = Integer.parseInt("fdcedfbd", 16);
it returns a numberFormatException
but if I do this then there will be no problem.
int tempI = 0xfdcedfbd
Can anybody help me with this? Thank you.

Answer 1

It sounds like the real problem here is that someone is giving you an 8-byte unsigned value represented as a 16-character hexadecimal string, and they want you to convert that into the actual 8 bytes. Are they going to do bit manipulation on it or something?
If you successfully convert it into a long and they then try to do math with it, they'll run into the sign/overflow problems everyone else has been talking about. If they just want to look at the individual bits then it would make sense to do the conversion into a long.
You could do something like this:
  private static final String highOrderDigits = "89abcdef";
  public long hexToLong(String hex) {
    // Only process exactly 16-digit strings.
    if (hex.length() != 16) {
      throw new IllegalArgumentException("Expect 16-digit hex value");
    boolean overflow = false;
    // Check for most significant digit > 7, which would
    // cause problem for parseLong.
    int highOrderIndex = highOrderDigits.indexOf(hex.charAt(0));
    if (highOrderIndex > -1) {
      // Clear most-significant bit and set overflow flag.
      overflow = true;
      hex = "" + highOrderIndex + hex.substring(1);
    long result = Long.parseLong(hex, 16);
    if (overflow) {
      // Manually set most significant bit in result.
      result |= 0x8000000000000000L;
    return result;
  }I more or less tested this and it looks to work.