type casting of byte and int

Question

I\'d a code snippet:

class AutoTypeCast{
    public static void main(String...args){
        int x=10;
        byte b=20;//no compilation error
        byte c=x;         


        

Answer 1

In this case x is initialized to 10, so there will be no data loss in the conversion from a 32-bit int to an 8-bit byte. But in general, when converting from int to byte, there can be data loss, so the rules of the Java language forbid assigning int values to a byte without a cast. This rule is designed to make it more difficult to write buggy code. By inserting the (byte) cast, you are effectively telling the compiler: "Yes, I have thought about the possibility of data loss, and it is not a problem here (or, that's actually what I want)."

Answer 2

When the compiler looks at the line

byte b=20;

it knows it's looking for a byte after b=. When it finds a constant numeric value it knows at compile time that it will definitely be in the range of byte so it will automatically cast it.

When is sees the line

byte c=x;

it's once again looking for a byte after c= but instead of finding a numeric constant, finds a variable that already has a defined type and, at compile time, can't be sure that it will be in the range of byte so you get an error.

Answer 3

X defined as integer, while narrowing there may be data loss, that is why compiler error. Refer jvm spec for conversions & promotions

Answer 4

20 always can be represented by a byte, while x, which according to the compiler can be any integer, may be too large to be represented by a byte.

Answer 5

20 is in -128..127 range, so it's value fits into byte.

Answer 6

Because compiler is not able to figure out the value of X at compile time. So it assume that X can contain value which is greater than byte range. If you make variable X as final then it will not give you compile time error.

        final int x=10;
        byte b=20;//no compilation error
        byte c=x;//no compilation error