Read two bytes into an integer?

问题

I have a byte[] that I've read from a file, and I want to get an int from two bytes in it. Here's an example:

byte[] bytes = new byte[] {(byte)0x00, (byte)0x2F, (byte)0x01, (byte)0x10, (byte)0x6F};
int value = bytes.getInt(2,4); //This method doesn't exist

This should make value equal to 0x0110, or 272 in decimal. But obviously, byte[].getInt() doesn't exist. How can I accomplish this task?

The above array is just an example. Actual values are unknown to me.

回答1:

You should just opt for the simple:

int val = ((bytes[2] & 0xff) << 8) | (bytes[3] & 0xff);

You could even write your own helper function getBytesAsWord (byte[] bytes, int start) to give you the functionality if you didn't want the calculations peppering your code but I think that would probably be overkill.

回答2:

Try:

public static int getInt(byte[] arr, int off) {
  return arr[off]<<8 &0xFF00 | arr[off+1]&0xFF;
} // end of getInt

Your question didn't indicate what the two args (2,4) meant. 2 and 4 don't make sense in your example as indices in the array to find ox01 and 0x10, I guessed you wanted to take two consecutive element, a common thing to do, so I used off and off+1 in my method.

You can't extend the byte[] class in java, so you can't have a method bytes.getInt, so I made a static method that uses the byte[] as the first arg.

The 'trick' to the method is that you bytes are 8 bit signed integers and values over 0x80 are negative and would be sign extended (ie 0xFFFFFF80 when used as an int). That is why the '&0xFF' masking is needed. the '<<8' shifts the more significant byte 8 bits left. The '|' combines the two values -- just as '+' would. The order of the operators is important because << has highest precedence, followed by & followed by | -- thus no parentheses are needed.

回答3:

Here's a nice simple reliable way.

    ByteBuffer byteBuffer = ByteBuffer.allocateDirect(4);
    // by choosing big endian, high order bytes must be put
    // to the buffer before low order bytes
    byteBuffer.order(ByteOrder.BIG_ENDIAN);
    // since ints are 4 bytes (32 bit), you need to put all 4, so put 0
    // for the high order bytes
    byteBuffer.put((byte)0x00);
    byteBuffer.put((byte)0x00);
    byteBuffer.put((byte)0x01);
    byteBuffer.put((byte)0x10);
    byteBuffer.flip();
    int result = byteBuffer.getInt();

回答4:

Alternatively, you could use:

int val = (bytes[2] << 8) + bytes[3]

回答5:

You can use ByteBuffer. It has the getInt method you are searching for and many other useful methods

回答6:

The Google Base16 class is from Guava-14.0.1.

new BigInteger(com.google.common.io.BaseEncoding.base16().encode(bytesParam),16).longValue();

来源：https://stackoverflow.com/questions/4768933/read-two-bytes-into-an-integer

标签

java

integer

byte

bytearray