my understanding says it should give a compile time error possible loss of precision. But it compiles And if we remove the final from i the as per my assumption it gives the error. Question is what is this final doing here to compile the code neatly in first case. Pinky
by definition -- when you have final int i = 10 -- the compiler knows without a doubt, that the value of i is going to be 10. All the time. So.... the compiler is also smart enough to know that the value 10 can fit into a byte without any loss of precision. BUT -- when you take away that final modifier -- the comipler can't guarantee what the value of the variable 'i' will be. So it throws a compile time error saying so and YOU have to take the responsibility (by casting it) to say -- yes I understand that there might be a loss of precision -- but I want you to do it anyway. ok.... so given that -- why does this give a "loss of precision" compile time error:
thanks that was a clear explaination. but does this mean if i leave the final var unassigned and later assign value using some variable cast will be required ? Pinky
OK Sant why won't this compile? should the compiler know that in this case i is going to be 30. If I is assigned at compile time then it should know the value of k and l. Even if k and l are not assigned a value, they get the default value of 0. Could you please explain why this fails to compile?
Hi Ryan, The compiler doesn't know that i is going to be 30. Even though i is final, k and l are not and can be changed. Hence i has to be evaluated at runtime. If you were to make k and l final, it would compile ok.
