我已使用接受的答案here“手动”从unicode转换为UTF-8代码单元。问题是我需要将生成的UTF-8包含在字节数组中。如何尽可能使用移位操作从十六进制到uft-8?
我已经拥有的代码如下:
public static void main(String[] args)
throws UnsupportedEncodingException, CharacterCodingException {
String st = "ñ";
for (int i = 0; i < st.length(); i++) {
int unicode = st.charAt(i);
codepointToUTF8(unicode);
}
}
public static byte[] codepointToUTF8(int codepoint) {
byte[] hb = codepointToHexa(codepoint);
byte[] binaryUtf8 = null;
if (codepoint <= 0x7F) {
binaryUtf8 = parseRange(hb, 8);
} else if (codepoint <= 0x7FF) {
binaryUtf8 = parseRange(hb, 16);
} else if (codepoint <= 0xFFFF) {
binaryUtf8 = parseRange(hb, 24);
} else if (codepoint <= 0x1FFFFF) {
binaryUtf8 = parseRange(hb, 32);
}
byte[] utf8Codeunits = new byte[hexStr.length()];
for (int i = 0; i < hexStr.length(); i++) {
utf8Codeunits[i] = (byte) hexStr.charAt(i);
System.out.println(utf8Codeunits[i]); // prints 99 51 98 49,
// which is the same as c3b1, the UTF-8 for ñ
}
return binaryUtf8;
}
public static byte[] codepointToHexa(int codepoint) {
int n = codepoint;
int m;
List<Byte> list = new ArrayList<>();
while (n >= 16) {
m = n % 16;
n = n / 16;
list.add((byte) m);
}
list.add((byte) n);
byte[] bytes = new byte[list.size()];
for (int i = list.size() - 1; i >= 0; i--) {
bytes[list.size() - i - 1] = list.get(i);
}
return bytes;
}
private static byte[] parseRange(byte[] hb, int length) {
byte[] binarybyte = new byte[length];
boolean[] filled = new boolean[length];
int index = 0;
if (length == 8) {
binarybyte[0] = 0;
filled[0] = true;
} else {
int cont = 0;
while (cont < length / 8) {
filled[index] = true;
binarybyte[index++] = 1;
cont++;
}
binarybyte[index] = 0;
filled[index] = true;
index = 8;
while (index < length) {
filled[index] = true;
binarybyte[index++] = 1;
binarybyte[index] = 0;
filled[index] = true;
index += 7;
}
}
byte[] hbbinary = convertHexaArrayToBinaryArray(hb);
int hbindex = hbbinary.length - 1;
for (int i = length - 1; i >= 0; i--) {
if (!filled[i] && hbindex >= 0) {
// we fill it and advance the iterator
binarybyte[i] = hbbinary[hbindex];
hbindex--;
filled[i] = true;
} else if (!filled[i]) {
binarybyte[i] = 0;
filled[i] = true;
}
}
return binarybyte;
}
private static byte[] convertHexaArrayToBinaryArray(byte[] hb) {
byte[] binaryArray = new byte[hb.length * 4];
String aux = "";
for (int i = 0; i < hb.length; i++) {
aux = Integer.toBinaryString(hb[i]);
int length = aux.length();
// toBinaryString doesn't return a 4 bit string, so we fill it with 0s
// if length is not a multiple of 4
while (length % 4 != 0) {
length++;
aux = "0" + aux;
}
for (int j = 0; j < aux.length(); j++) {
binaryArray[i * 4 + j] = (byte) (aux.charAt(j) - '0');
}
}
return binaryArray;
}
我不知道如何正确处理字节,所以我知道我所做的事情可能是错的。
答案 0 :(得分:2)
UTF-8按如下方式填充Unicode代码点:
0xxxxxxx
110xxxxx 10xxxxxx
1110xxxx 10xxxxxx 10xxxxxx
11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
... (max 6 bytes)
最右边的位数是该数字最不重要的位置。
static byte[] utf8(IntStream codePoints) {
final ByteArrayOutputStream baos = new ByteArrayOutputStream();
final byte[] cpBytes = new byte[6]; // IndexOutOfBounds for too large code points
codePoints.forEach((cp) -> {
if (cp < 0) {
throw new IllegalStateException("No negative code point allowed");
} else if (cp < 0x80) {
baos.write(cp);
} else {
int bi = 0;
int lastPrefix = 0xC0;
int lastMask = 0x1F;
for (;;) {
int b = 0x80 | (cp & 0x3F);
cpBytes[bi] = (byte)b;
++bi;
cp >>= 6;
if ((cp & ~lastMask) == 0) {
cpBytes[bi] = (byte) (lastPrefix | cp);
++bi;
break;
}
lastPrefix = 0x80 | (lastPrefix >> 1);
lastMask >>= 1;
}
while (bi > 0) {
--bi;
baos.write(cpBytes[bi]);
}
}
});
return baos.toByteArray();
}
除7位ASCII外,编码可以循环完成。