How do you specify a byte literal in Java?
You can use a byte literal in Java... sort of.
byte f = 0;
f = 0xa;
0xa
(int literal) gets automatically cast to byte. It's not a real byte literal (see JLS & comments below), but if it quacks like a duck, I call it a duck.
What you can't do is this:
void foo(byte a) {
...
}
foo( 0xa ); // will not compile
You have to cast as follows:
foo( (byte) 0xa );
But keep in mind that these will all compile, and they are using "byte literals":
void foo(byte a) {
...
}
byte f = 0;
foo( f = 0xa ); //compiles
foo( f = 'a' ); //compiles
foo( f = 1 ); //compiles
Of course this compiles too
foo( (byte) 1 ); //compiles
You have to cast, I'm afraid:
f((byte)0);
I believe that will perform the appropriate conversion at compile-time instead of execution time, so it's not actually going to cause performance penalties. It's just inconvenient :(
You cannot. A basic numeric constant is considered an integer (or long if followed by a "L"), so you must explicitly downcast it to a byte to pass it as a parameter. As far as I know there is no shortcut.