Question

byte b = 100 ; 

compiles without any errors, but

int i = 100 ; 
byte b = i ; 

throws an error. Why? Even when assigning 100 directly to b, we are assigning an int literal. So why did I get an error?

Answer 1

byte b = 100 ;

Here 100 is a compile time constant. And hence can be assigned to the byte.

int i = 100 ; 
// i = i + 100;  // It's allowed to add this statement later on, 
                 // if `i` is not declared final. And hence, the next assignment
                 // will clearly fail at runtime. So, compiler saves you from it
byte b = i ; 

Now in this case, since i is not declared final, so it is no more a compile time constant. And in that case, you can later on come and modify the value of i, in between the initialization of i, and assignment of i to byte, as in the above case. Which will certainly fail.
That is why compiler does not allow the assignment of i to byte type.

But, you can use an explicit casting for it to compile, which may of course crash at runtime. By doing an explicit cast, you tell the compiler that – “I know what I’m doing, just do it for me”. So, it will not bother about runtime behaviour of that casting, and will trust you that you are not doing anything wrong.

So, either you can declare your int i as final, or you need to do the casting: –

int i = 100 ;    // replace 100 with 130, and you will be surprised that it works
byte b = (byte)i ; 

So, when you use a value 130 and cast it to byte, you pass through the compiler, but will certainly crash at runtime. And this is the problem that the compiler was trying to avoid.

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Now let’s go back the first case: –

byte b = 128;

The above assignment will now fail to compile. Because even though the value 128 is a compile time constant, it is not big enough to fit in a byte, and that the compiler knows.