Java Automatic Type Promotion
In addition to assignments, there is another place also where a certain type conversions may occur, in expressions. To see why, consider this. In an expression, the precision required of an intermediate value will sometimes exceed the range of either the operand. For example, try the following expression :
byte a = 40; byte b = 50; byte c = 100; int d = a * b / c;
The output of the intermediate term, a * b easily exceeds the range of either of its byte operands.
To handle this type of problem, Java automatically promotes each byte, short, or char operand to int when evaluating an expression. It means that the subexpression a*b is performed using integers, not bytes. Thus, 2,000, the result of the intermediate expression, 50 * 40, is legal even though a and b both are specified as type byte.
As useful as the automatic promotions are, they can cause confusing compile-time errors also. For example, this seemingly correct code causes a problem :
byte b = 50; b = b * 2; // Error! Can't assign an int to a byte!
The code is attempting to store 50*2, a absolutely valid byte value, back into a byte variable. However, because the operands were automatically promoted to an int when the expression was evaluated, the result has also been promoted to int. Therefore, the result of the expression is now of type int, which cannot be assigned to a byte without the use of a cast. This is true even if, as in this case, the value being assigned would still fit in the target type.
In case where you understand the effect of overflow, you should use an explicit cast, such as
byte b = 50; b = (byte) (b * 2);
which yields the correct value of 100.
« Previous Tutorial Next Tutorial »