字符串或双精度字节

Question

I have a problem. I receive a double val=80.22. Then I split it:

String[] arr=String.valueOf(val).split("\\.");
int[] intArr=new int[2];
intArr[0]=Integer.parseInt(arr[0]); // 80
intArr[1]=Integer.parseInt(arr[1]); // 22

Now i need to put this values into a byte[] . But I don't need the value into the byte[] , I need it like this: 0x80 , 0x22.

What I have tried:

byte firstbyte = Byte.parseByte(arr[0], 16);
byte secondbyte = Byte.parseByte(arr[1], 16);

This works fine but only to the value 80, then I receive:

java.lang.NumberFormatException: Value out of range. Value:"80" Radix:16

But I don't get it. I need the number up to the value 100.

I have tried to write it into a string like string = 0x80 but I don't know how to put it into the byte[] then.

Thanks for your help!

Answer 1

Its hard for me to understand the question, but if you want to put a number up to the value 100 in a byte data type you just need to write:

    byte firstbyte = (byte)intArr[0];
    byte secondbyte = (byte)intArr[1];

Because 0x80 is an hexadecimal value and its worth to the decimal 128 which is too big for that data type.

Answer 2

Bytes are signed in Java. The highest value in a byte , in hexadecimal is 7F, so 80 is too big. Unfortunately there is no parseUnsignedByte .

The workaround is to use the type with the next higher range up, a short , then cast the result back to byte .

byte firstbyte = (byte) Short.parseShort(arr[0], 16);
byte secondbyte = (byte) Short.parseShort(arr[1], 16);

This will turn "80" into -128 , but the byte representation is correct: 1000 0000 , or 80 in hexadecimal.

Note that if the short value is out of the range of even an unsigned byte (at least 256, or "100" in hexadecimal), then this will silently truncate the value to the least significant 8 bits, ie 256 -> 0, 257 -> 1, etc.

Answer 3

byte max value is 2^7 - 1 = 127 in dec.

0x80 is hex representation of 128 in dec.

Hex counts from 0 to 9 then from A to F.

So, 80 in dec will be 0x50.

Simply avoid second parameter in Byte.parseByte() or use Byte.valueOf("80")