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Explain Arithmetic Promotion and Binary Operations?
An int divided through an int is an int, and a double divided by a double is a double, but what about an int divided by a double or a double divided by an int? When doing arithmetic on not like types Java tends to widen the types includes so as to prevent losing information. After all 3 * 54.2E18 will be a perfectly valid double but much too big for any int.
The primary rule is that if either of the variables in a binary operation (addition, multiplication, subtraction, addition, remainder) are doubles then Java treats both values as doubles. If neither value is a double but one is a float, then Java treats both values as floats. If neither is a float or a double but one is a long, then Java treats both values as longs. At last if there are no doubles, floats or longs, then Java treats both values as an int, even if there aren't any ints in the equation. Thus the result will be a double, float, long or int depending on the kinds of the arguments.
You should override the hashCode() and equals() functions from the Object class. The default implementation of the hashcode() and equals(), which are inherited from the java.
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