Your array is of
byte primitives, but you’re trying to call a method on them.
You don’t need to do anything explicit to convert a
byte to an
…since it’s not a downcast.
Note that the default behavior of
int conversion is to preserve the sign of the value (remember
byte is a signed type in Java). So for instance:
byte b1 = -100; int i1 = b1; System.out.println(i1); // -100
If you were thinking of the
byte as unsigned (156) rather than signed (-100), as of Java 8 there’s
byte b2 = -100; // Or `= (byte)156;` int = Byte.toUnsignedInt(b2); System.out.println(i2); // 156
Prior to Java 8, to get the equivalent value in the
int you’d need to mask off the sign bits:
byte b2 = -100; // Or `= (byte)156;` int i2 = (b2 & 0xFF); System.out.println(i2); // 156
Just for completeness #1: If you did want to use the various methods of
Byte for some reason (you don’t need to here), you could use a boxing conversion:
Byte b = rno; // Boxing conversion converts `byte` to `Byte` int i = b.intValue();
Byte b = new Byte(rno); int i = b.intValue();
But again, you don’t need that here.
Just for completeness #2: If it were a downcast (e.g., if you were trying to convert an
int to a
byte), all you need is a cast:
int i; byte b; i = 5; b = (byte)i;
This assures the compiler that you know it’s a downcast, so you don’t get the “Possible loss of precision” error.