Question

Consider this code:

public class ShortDivision {
    public static void main(String[] args) {
        short i = 2;
        short j = 1;
        short k = i/j;
    }
}

Compiling this produces the error

ShortDivision.java:5: possible loss of precision
found   : int
required: short
        short k = i/j;

because the type of the expression i/j is apparently int, and hence must be cast to short.

Why is the type of i/j not short?

Answer 1

From the Java spec:

5.6.2 Binary Numeric Promotion

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value of a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:

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.

For binary operations, small integer types are promoted to int and the result of the operation is int.

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EDIT: Why is it like that? The short answer is that Java copied this behavior from C. A longer answer might have to do with the fact that all modern machines do at least 32-bit native computations, and it might actually be harder for some machines to do 8-bit and 16-bit operations.

See also: OR-ing bytes in C# gives int