在Java中，为什么我的字节数组的元素包含2个字节？

Question

I am programming a little Java based Android app that receives ac based unsigned char[] byte array from a Bluetooth chip and streams it into a byte array.

The fact that it is Android and Bluetooth shouldn't matter though, it is just background. But, I am using minimum API 8 if that makes a difference.

The main code is:

InputStream is = socket.getInputStream();
DataInputStream dis = new DataInputStream(is);
byte[] buffer = new byte[1024];
bytesRead = dis.read(buffer, 0, 1024);

However, when I look at the actual contents of the buffer, I see this:

[0] 15 [0xf] [^O]   
[1] 15 [0xf] [^O]   
[2] 0 [0x0] [^@ (NUL)]  
[3] -119 [137] [0x89] [^É]  
[4] 2 [0x2] [^B]    
[5] 6 [0x6] [^F]    
[6] 26 [0x1a] [^Z]  
[7] -47 [209] [0xd1] [Ñ]    
[8] -1 [255] [0xff] [ÿ] 
[9] 104 [0x68] [h]  
[10]    -1 [255] [0xff] [ÿ] 
[11]    -46 [210] [0xd2] [Ò]    
[12]    -1 [255] [0xff] [ÿ] 
[13]    104 [0x68] [h]  
[14]    -19 [237] [0xed] [í]    
[15]    -128 [128] [0x80] [^À]  

The above is a copy from the eclipse Expression view set to show

1. Element [#]
2. Signed int value
3. If negative, unsigned int value [This value does not appear unless #2 is negative]
4. Hex value
5. ASCII value

My question is, if this is a byte array, why are some of the Hex values containing 2 bytes. Look at elements 6 through 14, each of them are of the form 0x1a, 0x12, 0xff, etc. The bytes 0 through 5 are all one byte (except for element 3).

I don't think this is an issue from the Bluetooth side because I see the actual code being made, and its an unsigned char[] array, each element is only one byte. Plus, I recall seeing something like this in a previous little project that involved taking data streams from an online API.

How can I ensure the Java array elements contain only 1 byte? And for that matter, I feel like I am not understanding something important about Java since this perplexes me -- how can Java allow a byte array to contain more than one byte per element?

Answer 1

You're confusing bytes and hexadecimal digits.

A byte contains 8 bits. An hexadecimal digit goes from 0 to F, and is thus only 4 bits (16 values, 16 = 2^4, thus 4 bits). A single byte is thus represented using two hexadecimal digits.

0xf is equivalent to 0x0f . The first 4 bits are all 0, and the last 4 bits are all 1.

Answer 2

A byte is 8 bits and so can hold 0 - (2^8 - 1) ie 0-255 (0x0 - 0xff). All your values are a single byte.