Narrowing Type Conversion: Why is assignment of int to a byte in a declaration allowed?

Question

This may sound too trivial for an intermediate Java programmer. But during my process of reviewing Java fundamentals, found a question:

Why is narrowing conversion like:

byte b = 13;

will be allowed while

int i = 13;
byte b = i;

will be complained by the compiler?

Answer 1

Because byte b = 13 ; is assignment of a constant. Its value is known at compile time, so the compiler can/should/will whine if assignment of the constant's value would result in overflow (try byte b = 123456789 ; and see what happens.)

Once you assign it to a variable, you're assigning the value of an expression, which, while it may well be invariant, the compiler doesn't know that. That expression might result in overflow and so the compiler whines.

Answer 2

From here:

Assignment conversion occurs when the value of an expression is assigned (§15.26) to a variable: the type of the expression must be converted to the type of the variable. Assignment contexts allow the use of an identity conversion (§5.1.1), a widening primitive conversion (§5.1.2), or a widening reference conversion (§5.1.4). In addition, a narrowing primitive conversion may be used if all of the following conditions are satisfied:

• The expression is a constant expression of type byte, short, char or int.
• The type of the variable is byte, short, or char.
• The value of the expression (which is known at compile time, because it is a constant expression) is representable in the type of the variable.

In your example all three conditions are satisfied, so the narrowing conversion is allowed.

P.S. I know the source I'm quoting is old, but this aspect of the language hasn't changed since.