Question

Let’s consider the following expressions in Java.

byte a = 32;
byte b = (byte) 250;
int i = a + b;

This is valid in Java even though the expression byte b = (byte) 250; is forced to assign the value 250 to b which is outside the range of the type byte. Therefore, b is assigned -6 and consequently i is assigned the value 26 through the statement int i = a + b;.

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The same thing is possible with short as follows.

short s1=(short) 567889999;

Although the specified value is outside the range of short, this statement is legal.

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The same thing is however wrong with higher data types such int, double, folat etc and hence, the following case is invalid and causes a compile-time error.

int z=2147483648;

This is illegal, since the range of int in Java is from -2,147,483,648 to 2147483647 which the above statement exceeds and issues a compile-time error. Why is such not wrong with byte and short data types in Java?

Answer 1

The difference is in the literal itself. You can do:

int z = (int) 2147483648L;

because now you’ve got a “long” literal. So basically, there are two steps involved:

• Parsing the literal to a valid value for that literal type (int in your earlier examples, long for the “big” one)
• Converting that value to the target type