Promotion in Java?

Question

The rules for promotion is "when operands are of different types, automatic binary numeric promotion occurs with the smaller operand type being converted to the larger". But the operands are of same type for example,

byte=byte+byte // Compile time error... found int..

So why is it so?

Answer 1

There's no + operator for byte. Instead, both operands are promoted to int, so you've got

byte = byte + byte
... becomes (widening to find + operator) ...
byte = int + int
... becomes (result of + operator) ...
byte = int

... which then fails because there's no implicit conversion from int to byte. You need to cast:

byte a = 1;
byte b = 2;

byte c = (byte) (a + b);

Here are the actual rules for numeric promotion, from section 5.6.2 of the JLS:

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

• If any of the operands is of a reference type, unboxing conversion (§5.1.8) is performed. Then:
• 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.

Answer 2

You were provided with correct answer about automatic promotion to 'int'.

There is one more note about that - compound assignment operators behave as they have an implicit type case. Example:

byte b1 = 1;
byte b2 = 2;
b1 = b1 + b2; // compilation fails
b1 += b2; // compilation successful