Java中的十六进制转整数 [英] Hexadecimal to Integer in Java
问题描述
我正在尝试将十六进制字符串转换为整数.字符串十六进制是从散列函数 (sha-1) 计算出来的.我收到此错误:java.lang.NumberFormatException.我猜它不喜欢十六进制的字符串表示.我怎样才能做到这一点.这是我的代码:
I am trying to convert a String hexadecimal to an integer. The string hexadecimal was calculated from a hash function (sha-1). I get this error : java.lang.NumberFormatException. I guess it doesn't like the String representation of the hexadecimal. How can I achieve that. Here is my code :
public Integer calculateHash(String uuid) {
try {
MessageDigest digest = MessageDigest.getInstance("SHA1");
digest.update(uuid.getBytes());
byte[] output = digest.digest();
String hex = hexToString(output);
Integer i = Integer.parseInt(hex,16);
return i;
} catch (NoSuchAlgorithmException e) {
System.out.println("SHA1 not implemented in this system");
}
return null;
}
private String hexToString(byte[] output) {
char hexDigit[] = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
'A', 'B', 'C', 'D', 'E', 'F' };
StringBuffer buf = new StringBuffer();
for (int j = 0; j < output.length; j++) {
buf.append(hexDigit[(output[j] >> 4) & 0x0f]);
buf.append(hexDigit[output[j] & 0x0f]);
}
return buf.toString();
}
例如，当我传递这个字符串时:_DTOWsHJbEeC6VuzWPawcLA，他的哈希值是:0xC934E5D372B2AB6D0A50B9F0341A00ED029BDC15
For example, when I pass this string : _DTOWsHJbEeC6VuzWPawcLA, his hash his : 0xC934E5D372B2AB6D0A50B9F0341A00ED029BDC15
但我得到:java.lang.NumberFormatException:对于输入字符串:0xC934E5D372B2AB6D0A50B9F0341A00ED029BDC15"
But i get : java.lang.NumberFormatException: For input string: "0xC934E5D372B2AB6D0A50B9F0341A00ED029BDC15"
我真的需要这样做.我有一组由它们的 UUID 标识的元素，它们是字符串.我将不得不存储这些元素，但我的限制是使用整数作为它们的 id.这就是为什么我计算给定参数的哈希值，然后将其转换为 int.也许我做错了，但有人可以给我一个建议来正确地实现它！
I really need to do this. I have a collection of elements identified by their UUID which are string. I will have to store those elements but my restrictions is to use an integer as their id. It is why I calculate the hash of the parameter given and then I convert to an int. Maybe I am doing this wrong but can someone gives me an advice to achieve that correctly!!
感谢您的帮助！！
推荐答案
为什么不使用 java 功能呢:
Why do you not use the java functionality for that:
如果您的数字很小(比您的小)，您可以使用:Integer.parseInt(hex, 16) 将十六进制 - 字符串转换为整数.
If your numbers are small (smaller than yours) you could use: Integer.parseInt(hex, 16) to convert a Hex - String into an integer.
String hex = "ff"
int value = Integer.parseInt(hex, 16);
对于像你这样的大数字，使用 public BigInteger(String val, int radix)
For big numbers like yours, use public BigInteger(String val, int radix)
BigInteger value = new BigInteger(hex, 16);
@见JavaDoc:
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