JAVA：从字符串中获取 UTF-8 十六进制值？
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JAVA: get UTF-8 Hex values from a string?
提问by thedp
I would like to be able to convert a raw UTF-8 string to a Hex string. In the example below I've created a sample UTF-8 string containing 2 letters. Then I'm trying to get the Hex values but it gives me negative values.
我希望能够将原始 UTF-8 字符串转换为十六进制字符串。在下面的示例中，我创建了一个包含 2 个字母的示例 UTF-8 字符串。然后我试图获得十六进制值，但它给了我负值。
How can I make it give me 05D0and 05D1
我怎样才能让它给我05D0和05D1
String a = "\u05D0\u05D1";
byte[] xxx = a.getBytes("UTF-8");
for (byte x : xxx) {
System.out.println(Integer.toHexString(x));
}
Thank you.
谢谢你。
采纳答案by ataylor
Don't convert to an encoding like UTF-8 if you want the code point. Use Character.codePointAt.
如果需要代码点，请不要转换为 UTF-8 之类的编码。使用Character.codePointAt。
For example:
例如：
Character.codePointAt("\u05D0\u05D1", 0) // returns 1488, or 0x5d0
回答by Malcolm
Negative values occur because the range of byteis from -128 to 127. The following code will produce positive values:
出现负值byte是因为 的范围是从 -128 到 127。以下代码将产生正值：
String a = "\u05D0\u05D1";
byte[] xxx = a.getBytes("UTF-8");
for (byte x : xxx) {
System.out.println(Integer.toHexString(x & 0xFF));
}
The main difference is that it outputs x & 0xFFinstead of just x, this operation converts byteto int, dropping the sign.
主要区别在于它输出x & 0xFF而不是仅输出x，此操作转换byte为int，丢弃符号。
