Operators

[Andreas Svenkson]

Evening guys.

I'm a bit tired due to studying all day, and right now nothing makes sense to me. Could someone explain this?:

public class TestCompound {
	public static void main(String...args) {
		
		int k = 1, l = 2;
		byte b = 1;
		b *= k * l;
		System.out.println(b);
}
}

I thought that more or less any operation between 2 "integers" would result in an integer, assuming none of the integers are a long, in which case it all turns into a long. So multiplying 2 bytes like here (k and l) would result in an integer, being unassignable to our byte b.
This is also the case, if we do not use a compound operator: *= and only use an assignment operator =.

So, why does b *= k*l; compile when b = k*l; does not ?

Hope that was clear...

// Andreas

[Stephan van Hulst]

Compound assignment operators perform silent casts.

a @= b

means the same as

a = (T) (a @ b)

where @ is an operator that has a compound assignment counterpart, and T is the type of a.

[Andreas Svenkson]

Thanks mate, I did not know that.

// Andreas