我想在Java中将字符数组转换为字节数组。 有哪些方法可以进行此转换?
不创建String
对象进行转换:
import java.nio.CharBuffer;
import java.nio.ByteBuffer;
import java.util.Arrays;
byte[] toBytes(char[] chars) {
CharBuffer charBuffer = CharBuffer.wrap(chars);
ByteBuffer byteBuffer = Charset.forName("UTF-8").encode(charBuffer);
byte[] bytes = Arrays.copyOfRange(byteBuffer.array(),
byteBuffer.position(), byteBuffer.limit());
Arrays.fill(byteBuffer.array(), (byte) 0); // clear sensitive data
return bytes;
}
使用方法:
char[] chars = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'};
byte[] bytes = toBytes(chars);
/* do something with chars/bytes */
Arrays.fill(chars, '\u0000'); // clear sensitive data
Arrays.fill(bytes, (byte) 0); // clear sensitive data
解决方案灵感来源于 Swing 推荐将密码存储在 char[] 中。(参见 为什么 char[] 被优先于 String 用于密码存储?)
记住不要将敏感数据写入日志,并确保 JVM 不会持有任何对其的引用。
此方法仅用于安全方面的考虑。如果数据不敏感,最好使用 String.getBytes
。
val xs: Array[Char] = "A ß € 嗨 ".toArray
val len = xs.length
val ys: Array[Byte] = new Array(3 * len) // worst case
var i = 0; var j = 0 // i for chars; j for bytes
while (i < len) { // fill ys with bytes
val c = xs(i)
if (c < 0x80) {
ys(j) = c.toByte
i = i + 1
j = j + 1
} else if (c < 0x800) {
ys(j) = (0xc0 | (c >> 6)).toByte
ys(j + 1) = (0x80 | (c & 0x3f)).toByte
i = i + 1
j = j + 2
} else if (Character.isHighSurrogate(c)) {
if (len - i < 2) throw new Exception("overflow")
val d = xs(i + 1)
val uc: Int =
if (Character.isLowSurrogate(d)) {
Character.toCodePoint(c, d)
} else {
throw new Exception("malformed")
}
ys(j) = (0xf0 | ((uc >> 18))).toByte
ys(j + 1) = (0x80 | ((uc >> 12) & 0x3f)).toByte
ys(j + 2) = (0x80 | ((uc >> 6) & 0x3f)).toByte
ys(j + 3) = (0x80 | (uc & 0x3f)).toByte
i = i + 2 // 2 chars
j = j + 4
} else if (Character.isLowSurrogate(c)) {
throw new Exception("malformed")
} else {
ys(j) = (0xe0 | (c >> 12)).toByte
ys(j + 1) = (0x80 | ((c >> 6) & 0x3f)).toByte
ys(j + 2) = (0x80 | (c & 0x3f)).toByte
i = i + 1
j = j + 3
}
}
// check
println(new String(ys, 0, j, "UTF-8"))
char[] ch = ?
new String(ch).getBytes();
或者,要获取非默认字符集:
new String(ch).getBytes("UTF-8");
更新:自Java 7起:
new String(ch).getBytes(StandardCharsets.UTF_8);
Andrey的答案(在撰写本文时获得最高投票)略有误。虽然我想以评论的形式添加,但我声望不够。
在Andrey的答案中:
char[] chars = {'c', 'h', 'a', 'r', 's'}
byte[] bytes = Charset.forName("UTF-8").encode(CharBuffer.wrap(chars)).array();
调用array()函数可能无法返回预期的值,例如:
char[] c = "aaaaaaaaaa".toCharArray();
System.out.println(Arrays.toString(Charset.forName("UTF-8").encode(CharBuffer.wrap(c)).array()));
输出:
[97, 97, 97, 97, 97, 97, 97, 97, 97, 97, 0]
可以看到已经添加了一个零字节。为避免这种情况,使用以下方法:
char[] c = "aaaaaaaaaa".toCharArray();
ByteBuffer bb = Charset.forName("UTF-8").encode(CharBuffer.wrap(c));
byte[] b = new byte[bb.remaining()];
bb.get(b);
System.out.println(Arrays.toString(b));
输出:
[97, 97, 97, 97, 97, 97, 97, 97, 97, 97]
正如答案所提到的使用密码,清空支持ByteBuffer(通过array()函数访问)的数组可能是值得的:
ByteBuffer bb = Charset.forName("UTF-8").encode(CharBuffer.wrap(c));
byte[] b = new byte[bb.remaining()];
bb.get(b);
blankOutByteArray(bb.array());
System.out.println(Arrays.toString(b));
toBytes
函数中,buffer.array()
仍需要被覆盖,目前只有复制。 - djsuthoprivate static byte[] charArrayToByteArray(char[] c_array) {
byte[] b_array = new byte[c_array.length];
for(int i= 0; i < c_array.length; i++) {
b_array[i] = (byte)(0xFF & (int)c_array[i]);
}
return b_array;
}
// original byte[]
byte[] pattern = null;
char[] arr = new char[pattern.length * 2];
ByteBuffer wrapper = ByteBuffer.wrap(pattern);
wrapper.position(0);
int i = 0;
while(wrapper.hasRemaining()) {
char character = wrapper.remaining() < 2 ? ((char) (((int) wrapper.get()) << 8)) : wrapper.getChar();
arr[i++] = character;
}
public byte[] toBytes(char[] data) {
byte[] toRet = new byte[data.length];
for(int i = 0; i < toRet.length; i++) {
toRet[i] = (byte) data[i];
}
return toRet;
}
希望这能有所帮助