byte b1 = (byte) 129;
String s1 = String.format("%8s", Integer.toBinaryString(b1 & 0xFF)).replace(' ', '0');
System.out.println(s1); // 10000001
byte b2 = (byte) 2;
String s2 = String.format("%8s", Integer.toBinaryString(b2 & 0xFF)).replace(' ', '0');
System.out.println(s2); // 00000010
演示。
我用过这个。与其他答案类似的想法，但在任何地方都没有看到确切的方法:)
System.out.println(Integer.toBinaryString((b & 0xFF) + 0x100).substring(1));
0xFF是 255，或11111111（无符号字节的最大值）。
0x100是 256，或100000000
将&字节向上转换为整数。那时，它可以是0- 255(00000000到11111111，我排除了前 24 位)。+ 0x100并.substring(1)确保会有前导零。
与 João Silva 的答案相比，我对其进行了计时，这比João Silva 的答案快了 10 倍以上。http://ideone.com/22DDK1我没有包括 Pshemo 的答案，因为它没有正确填充。
这是你想要的？
从字符串转换为字节
byte b = (byte)(int)Integer.valueOf("10000010", 2);
System.out.println(b);// output -> -126
从字节转换为字符串
System.out.println(Integer.toBinaryString((b+256)%256));// output -> "10000010"
或者正如 João Silva 在他的评论中所说的添加前导0，我们可以将字符串格式化为长度 8，并将生成的前导空格替换为零，所以在字符串的情况下，" 1010"我们将得到"00001010"
System.out.println(String.format("%8s", Integer.toBinaryString((b + 256) % 256))
.replace(' ', '0'));
您可以检查字节上的每个位，然后将 0 或 1 附加到字符串。这是我为测试编写的一个小辅助方法：
public static String byteToString(byte b) {
byte[] masks = { -128, 64, 32, 16, 8, 4, 2, 1 };
StringBuilder builder = new StringBuilder();
for (byte m : masks) {
if ((b & m) == m) {
builder.append('1');
} else {
builder.append('0');
}
}
return builder.toString();
}
获取字节的每一位并转换为字符串。假设字节有 8 位，我们可以通过位移动一个一个地得到它们。例如，我们将 6 位字节的第二位向右移动，将 8 位的第二位移到最后一位，然后用 0x0001 与 (&) 清除前面的位。
public static String getByteBinaryString(byte b) {
StringBuilder sb = new StringBuilder();
for (int i = 7; i >= 0; --i) {
sb.append(b >>> i & 1);
}
return sb.toString();
}
此代码将演示如何将 java int 拆分为其 4 个连续字节。然后，与低级字节/位询问相比，我们可以使用 Java 方法检查每个字节。
这是运行以下代码时的预期输出：
[Input] Integer value: 8549658
Integer.toBinaryString: 100000100111010100011010
Integer.toHexString: 82751a
Integer.bitCount: 10
Byte 4th Hex Str: 0
Byte 3rd Hex Str: 820000
Byte 2nd Hex Str: 7500
Byte 1st Hex Str: 1a
(1st + 2nd + 3rd + 4th (int(s)) as Integer.toHexString: 82751a
(1st + 2nd + 3rd + 4th (int(s)) == Integer.toHexString): true
Individual bits for each byte in a 4 byte int:
00000000 10000010 01110101 00011010
这是要运行的代码：
public class BitsSetCount
{
public static void main(String[] args)
{
int send = 8549658;
System.out.println( "[Input] Integer value: " + send + "\n" );
BitsSetCount.countBits( send );
}
private static void countBits(int i)
{
System.out.println( "Integer.toBinaryString: " + Integer.toBinaryString(i) );
System.out.println( "Integer.toHexString: " + Integer.toHexString(i) );
System.out.println( "Integer.bitCount: "+ Integer.bitCount(i) );
int d = i & 0xff000000;
int c = i & 0xff0000;
int b = i & 0xff00;
int a = i & 0xff;
System.out.println( "\nByte 4th Hex Str: " + Integer.toHexString(d) );
System.out.println( "Byte 3rd Hex Str: " + Integer.toHexString(c) );
System.out.println( "Byte 2nd Hex Str: " + Integer.toHexString(b) );
System.out.println( "Byte 1st Hex Str: " + Integer.toHexString(a) );
int all = a+b+c+d;
System.out.println( "\n(1st + 2nd + 3rd + 4th (int(s)) as Integer.toHexString: " + Integer.toHexString(all) );
System.out.println("(1st + 2nd + 3rd + 4th (int(s)) == Integer.toHexString): " +
Integer.toHexString(all).equals(Integer.toHexString(i) ) );
System.out.println( "\nIndividual bits for each byte in a 4 byte int:");
/*
* Because we are sending the MSF bytes to a method
* which will work on a single byte and print some
* bits we are generalising the MSF bytes
* by making them all the same in terms of their position
* purely for the purpose of printing or analysis
*/
System.out.print(
getBits( (byte) (d >> 24) ) + " " +
getBits( (byte) (c >> 16) ) + " " +
getBits( (byte) (b >> 8) ) + " " +
getBits( (byte) (a >> 0) )
);
}
private static String getBits( byte inByte )
{
// Go through each bit with a mask
StringBuilder builder = new StringBuilder();
for ( int j = 0; j < 8; j++ )
{
// Shift each bit by 1 starting at zero shift
byte tmp = (byte) ( inByte >> j );
// Check byte with mask 00000001 for LSB
int expect1 = tmp & 0x01;
builder.append(expect1);
}
return ( builder.reverse().toString() );
}
}
对不起，我知道这有点晚了......但我有一个更简单的方法......二进制字符串：
//Add 128 to get a value from 0 - 255
String bs = Integer.toBinaryString(data[i]+128);
bs = getCorrectBits(bs, 8);
getCorrectBits 方法：
private static String getCorrectBits(String bitStr, int max){
//Create a temp string to add all the zeros
StringBuilder sb = new StringBuilder();
for(int i = 0; i < (max - bitStr.length()); i ++){
sb.append("0");
}
return sb.toString()+ bitStr;
}
Integer.toBinaryString((byteValue & 0xFF) + 256).substring(1)
您可以像下面的示例一样使用BigInteger，尤其是如果您有256 位或更长：
String string = "10000010";
BigInteger biStr = new BigInteger(string, 2);
System.out.println("binary: " + biStr.toString(2));
System.out.println("hex: " + biStr.toString(16));
System.out.println("dec: " + biStr.toString(10));
另一个接受字节的例子：
String string = "The girl on the red dress.";
byte[] byteString = string.getBytes(Charset.forName("UTF-8"));
System.out.println("[Input String]: " + string);
System.out.println("[Encoded String UTF-8]: " + byteString);
BigInteger biStr = new BigInteger(byteString);
System.out.println("binary: " + biStr.toString(2)); // binary
System.out.println("hex: " + biStr.toString(16)); // hex or base 16
System.out.println("dec: " + biStr.toString(10)); // this is base 10
结果：
[Input String]: The girl on the red dress.
[Encoded String UTF-8]: [B@70dea4e
binary: 101010001101000011001010010000001100111011010010111001001101100001000000110111101101110001000000111010001101000011001010010000001110010011001010110010000100000011001000111001001100101011100110111001100101110
hex: 546865206769726c206f6e20746865207265642064726573732e
您还可以将二进制转换为字节格式
try {
System.out.println("binary to byte: " + biStr.toString(2).getBytes("UTF-8"));
} catch (UnsupportedEncodingException e) {e.printStackTrace();}
注意： 对于二进制格式的字符串格式，您可以使用以下示例
String.format("%256s", biStr.toString(2).replace(' ', '0')); // this is for the 256 bit formatting
String byteToBinaryString(byte b){
StringBuilder binaryStringBuilder = new StringBuilder();
for(int i = 0; i < 8; i++)
binaryStringBuilder.append(((0x80 >>> i) & b) == 0? '0':'1');
return binaryStringBuilder.toString();
}
一个简单的答案可能是：
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0})); // 0
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1})); // 1
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0})); // 256
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0})); // 65536
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0})); // 16777216
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0})); // 4294967296
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0})); // 1099511627776
System.out.println(new BigInteger(new byte[]{0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0})); // 281474976710656
System.out.println(new BigInteger(new byte[]{0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0})); // 72057594037927936
System.out.println(new BigInteger(new byte[]{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0})); // 18446744073709551616
System.out.println(new BigInteger(new byte[]{0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0})); // 4722366482869645213696
System.out.println(new BigInteger(new byte[]{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0})); // 1208925819614629174706176
System.out.println(Long.MAX_VALUE); // 9223372036854775807
我们都知道 Java 不提供类似 unsigned 关键字的任何东西。此外，byte根据 Java 规范的原语表示 和 之间的−128值127。例如，如果 abyte是Java 将把第一个解释cast为and使用符号扩展。intbitsign
127为其二进制字符串表示？没有什么能阻止您将 abyte简单地视为 8 位并将这些位解释为 和 之间的0值255。此外，您需要记住，您无法将自己的解释强加于其他人的方法。如果一个方法接受 a byte，那么该方法接受一个介于−128and之间的值，127除非另有明确说明。
所以解决这个问题的最好方法是通过调用方法或将其转换为原始byte值来将值转换为值。这里有一个例子： intByte.toUnsignedInt()int(int) signedByte & 0xFF
public class BinaryOperations
{
public static void main(String[] args)
{
byte forbiddenZeroBit = (byte) 0x80;
buffer[0] = (byte) (forbiddenZeroBit & 0xFF);
buffer[1] = (byte) ((forbiddenZeroBit | (49 << 1)) & 0xFF);
buffer[2] = (byte) 96;
buffer[3] = (byte) 234;
System.out.println("8-bit header:");
printBynary(buffer);
}
public static void printBuffer(byte[] buffer)
{
for (byte num : buffer) {
printBynary(num);
}
}
public static void printBynary(byte num)
{
int aux = Byte.toUnsignedInt(num);
// int aux = (int) num & 0xFF;
String binary = String.format("%8s', Integer.toBinaryString(aux)).replace(' ', '0');
System.out.println(binary);
}
}
8-bit header:
10000000
11100010
01100000
11101010
对于那些需要将字节大量转换为二进制字符串的人来说，这只是另一个提示：使用查找表而不是一直使用那些字符串操作。这比一遍又一遍地调用转换函数要快得多
public class ByteConverterUtil {
private static final String[] LOOKUP_TABLE = IntStream.range(0, Byte.MAX_VALUE - Byte.MIN_VALUE + 1)
.mapToObj(intValue -> Integer.toBinaryString(intValue + 0x100).substring(1))
.toArray(String[]::new);
public static String convertByte(final byte byteValue) {
return LOOKUP_TABLE[Byte.toUnsignedInt(byteValue)];
}
public static void main(String[] args){
System.out.println(convertByte((byte)0)); //00000000
System.out.println(convertByte((byte)2)); //00000010
System.out.println(convertByte((byte)129)); //10000001
System.out.println(convertByte((byte)255)); //11111111
}
}
只是在这里猜测，但如果你有一个 Byte 那么你不能简单地在对象上调用 toString() 来获取值吗？或者，使用 byteValue()查看api ？