Question

As i understand, when you cast between 2 of these types with an arithmetic operation in Java, for example double + int, the result will be as the bigger type (meaning in this example, the result will be double). What happens when you make an arithmetic operation on 2 types with the same size? what will int + float and long + double give? since int and float are 4 bytes each, and long and double are 8 bytes.

Answer 1

This is all specified by the binary numeric promotion rules in the JLS. From http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2:

When an operator applies binary numeric promotion to a pair of
operands, each of which must denote a value that is convertible to a
numeric type, the following rules apply, in order:

1. If any operand is of a reference type, it is subjected to unboxing
   conversion (§5.1.8).

2. Widening primitive conversion (§5.1.2) is applied to convert either or
   both operands as specified by the following rules:

   • If either operand is of type double, the other is converted to double.

   • Otherwise, if either operand is of type float, the other is converted
     to float.

   • Otherwise, if either operand is of type long, the other is converted
     to long.

   • Otherwise, both operands are converted to type int.

After the type conversion, if any, value set conversion (§5.1.13) is
applied to each operand.

Binary numeric promotion is performed on the operands of certain operators:

• The multiplicative operators *, / and % (§15.17)

• The addition and subtraction operators for numeric types + and –
  (§15.18.2)

• The numerical comparison operators <, <=, >, and >= (§15.20.1)

• The numerical equality operators == and != (§15.21.1)

• The integer bitwise operators &, ^, and | (§15.22.1)

• In certain cases, the conditional operator ? : (§15.25)

(“value set conversion” is about mapping between floating-point representations.)