Java 如何将整数拆分为 2 字节二进制文件？
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How do I split an integer into 2 byte binary?
提问by Kevin Boyd
Given
给定的
private int width = 400;
private byte [] data = new byte [2];
I want to split the integer "width" into two bytes and load data[0] with the high byte and data[1] with the low byte.
我想将整数“宽度”分成两个字节，并用高字节加载 data[0]，用低字节加载 data[1]。
That is binary value of 400 = 1 1001 0000 so data[0] should contain 0000 0001 and data[1] should contain 1001 0000
那是 400 = 1 1001 0000 的二进制值所以 data[0] 应该包含 0000 0001 并且 data[1] 应该包含 1001 0000
采纳答案by Stephan202
Using simple bitwise operations:
使用简单的按位运算：
data[0] = (byte) (width & 0xFF);
data[1] = (byte) ((width >> 8) & 0xFF);
How it works:
这个怎么运作：
& 0xFFmasks all but the lowest eight bits.>> 8discards the lowest 8 bits by moving all bits 8 places to the right.- The cast to byte is necessary because these bitwise operations work on an
intand return anint, which is a bigger data type thanbyte. The case is safe, since all non-zero bits will fit in thebyte. For more information, see Conversions and Promotions.
& 0xFF屏蔽除最低八位之外的所有位。>> 8通过将所有位向右移动 8 位来丢弃最低 8 位。- 强制转换为字节是必要的，因为这些按位运算适用于 an
int并返回 anint，这是比byte. 这种情况是安全的，因为所有非零位都适合byte. 有关更多信息，请参阅转化和促销。
Edit:Taylor Lcorrectly remarks that though >>works in this case, it may yield incorrect results if you generalize this code to fourbytes (since in Java an intis 32 bits). In that case, it's better to use >>>instead of >>. For more information, see the Java tutorial on Bitwise and Bit Shift Operators.
编辑：Taylor L正确地指出，虽然>>在这种情况下有效，但如果将此代码概括为四个字节（因为在 Java 中int是 32 位），它可能会产生不正确的结果。在这种情况下，最好使用>>>代替>>。有关详细信息，请参阅有关按位和位移运算符的Java 教程。
回答by Dmitry
Integer is 32 bits (=4 bytes) in java, you know?
Java 中的整数是 32 位（=4 字节），你知道吗？
width & 0xff will give you the first byte, width & 0xff00 >> 8 will give you the second, etc.
width & 0xff 会给你第一个字节， width & 0xff00 >> 8 会给你第二个，依此类推。
回答by Rod Hyde
To get the high byte, shift right by 8 bits then mask off the top bytes. Similarly, to get the low byte just mask off the top bytes.
要获得高字节，请右移 8 位然后屏蔽高字节。同样，要获得低字节，只需屏蔽高字节。
data[0] = (width >> 8) & 0xff;
data[1] = width & 0xff;
回答by alltom
int width = 400;
byte [] data = new byte [2];
data[0] = (byte) ((width & 0xFF00) >> 8);
data[1] = (byte) (width & 0xFF);
for(int b = 0; b < 2; b++) {
System.out.println("printing byte " + b);
for(int i = 7; i >= 0; i--) {
System.out.println(data[b] & 1);
data[b] = (byte) (data[b] >> 1);
}
}
回答by Taylor Leese
This should do what you want for a 4 byte int. Note, it stores the low byte at offset 0. I'll leave it as an exercise to the reader to order them as needed.
这应该为 4 字节 int 做你想要的。请注意，它将低字节存储在偏移量 0 处。我将把它留给读者作为练习，以便根据需要对它们进行排序。
public static byte[] intToBytes(int x) {
byte[] bytes = new byte[4];
for (int i = 0; x != 0; i++, x >>>= 8) {
bytes[i] = (byte) (x & 0xFF);
}
return bytes;
}
回答by starblue
For converting two bytes the cleanest solution is
为了转换两个字节，最干净的解决方案是
data[0] = (byte) width;
data[1] = (byte) (width >>> 8);
For converting an integer to four bytes the code would be
要将整数转换为四个字节，代码将是
data[0] = (byte) width;
data[1] = (byte) (width >>> 8);
data[2] = (byte) (width >>> 16);
data[3] = (byte) (width >>> 24);
It doesn't matter whether >> or >>> is used for shifting, any one bits created by sign extension will not end up in the resulting bytes.
无论是 >> 还是 >>> 用于移位，符号扩展创建的任何一位都不会在结果字节中结束。
See also this answer.
另请参阅此答案。
回答by Peter Lawrey
I suggest you have a look at the source for HeapByteBuffer. It has the conversion code for all primitive data types. (In fact you could just use a ByteBuffer ;)
我建议您查看 HeapByteBuffer 的源代码。它具有所有原始数据类型的转换代码。（事实上你可以只使用一个 ByteBuffer ;）
