Character arithmetic in Java

Question

While playing around I met something, that seems strange to me:

The following isn't valid Java code:

char x = 'A';
x = x + 1;    //possible loss of precision

because one of the operands ist an integer and so the other operand is converted to integer. The result couldn't be assigned to a character variable ... while

char x = 'A';
x += 1;

is valid, because the resulting integer is - automatically - converted to character.

So far so good. This seems clear to me but ... why is the following valid Java code?

char x;
x = 'A' + 1;

Answer 1

Because

'A' + 1

is a constant expression. It is known at compile time that the result will fit in a char.

Whereas

'A' + 787282;

will not fit in a char and will therefore cause a compilation error.

Answer 2

It is valid because it is a compile time constant expression. Had it been

char x;
char y = 'A';
x =  y + 1;

The compiler will give you a compile time error, because now it is not a compile time constant expression. But if you will make the variable y as final the expression will turn again into compile time constant , thus code below will compile.

char x;
final char y = 'A';
x =  y + 1;

Moral of the story is that when you assign integer to a char , the compiler will allow it as long as it is compiler time constant and it should fit in the range of the char.