A much simpler approach is to cast it:
int number = 204;
byte b = (byte)number;
System.out.println(b);
Prints -52
.
質問
Long story short, I'm reading some integer values from one file, then I need to store them in a byte array, for later writing to another file.
For example:
int number = 204;
Byte test = new Byte(Integer.toString(number));
This code throws:
java.lang.NumberFormatException: Value out of range. Value:"204" Radix:10
The problem here is that a byte can only store from -127 - 128, so obviously that number is far too high. What I need to do is have the number signed, which is the value -52, which will fit into the byte. However, I'm unsure how to accomplish this.
Can anyone advise?
Thanks
解決
A much simpler approach is to cast it:
int number = 204;
byte b = (byte)number;
System.out.println(b);
Prints -52
.
他のヒント
You can cast it:
Byte test = (byte) number;
Use this:
byte b = (byte) ((0xFF) & number);