Byte incrementing operation in Java [duplicate]

Answer 1

The difference is that there is an implicit casting in the ++ operator from int to byte, whereas, you would have to do that explicitly in case if you use b = b + 1

b = b + 1;  // Will not compile. Cannot cast from int to byte

You need an explicit cast:

b = (byte) (b + 1);

Whereas b++ will work fine. The ++ operator automatically casts the value b + 1, which is an int to a byte.

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This is clearly listed in JLS - §15.26.2 Compound Assignment Operators : -

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once

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Please note that operation b + 1 will give you a result of type int. So, that's why you need an explicit cast in your second assignment.

Answer 2

b++; and b=b+1; are equivalent and will result in the same bytecode.

b will be equal to 3 before the main method is finished.

Edit: actually they are not equivalent: b=b+1; is wrong and should be b=(byte) (b+1); [cast to a byte, otherwise it is an int]

Answer 3

The variable will be incremented twice

Answer 4

What happens? Actually b = b + 1 won't compile.

You must explicitly convert it to byte, because b + 1 evaluates to int. And it is not guaranteed that an int can fit into a byte.

b = (byte)(b + 1);