char and byte with final access modifier - java

byte, char, java, primitive, scjp

Solution

Because qualifying it with `final` makes the variable a constant variable, which is a constant expression. So

final byte b = 1;
char c = 2;

c = b; // line 2

actually becomes

final byte b = 1;
char c = 2;

c = 1;

And the compiler has a guarantee that the value `1` can fit in a `char` variable.

With a non constant `byte` variable, there is no such guarantee. `byte` is signed, `char` is unsigned.

Problem

Please take a look at below example i cant understand the relation between char and byte ``` byte b = 1; char c = 2; c = b; // line 1 ``` Give me compilation Error because c is type of `char` and b is type of `byte` so casting is must in such condition but now the tweest here is when i run below code ``` final byte b = 1; char c = 2; c = b; // line 2 ``` line 2 compile successfully it doesn't need any casting at all so my question is why `char` c behave different when i use final access modifier with `byte`

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