Java Casting 对我的误解

Question

I am new to java, and I don't understand the differences between these two:

        byte myByte = 100;
        short myShort = 5000 ;
        int myInt = 2_000_150_000;

        long myLong = (long)(50_000 + 10*(long)(myByte + myShort + myInt));
        System.out.println("test1:" + myLong);

        long myLongWrong = (long)(50_000 + 10*(myByte + myShort + myInt));
        System.out.println("test2:" + myLongWrong);

OUTPUT:

    test1:20001601000
    test2:-1473235480

I know whenever I got var and arithmetic I need to do a casting with (long) but why I need to do it outside the main () too?

THE SAME THINK FOR short work differently

    short myShortTest = (short)(50_000 + 10*(short)(myByte + myInt +myShort));
    short myShortTest2 = (short)(50_000 + 10*(myByte + myInt +myShort));
    System.out.println(myShortTest);
    System.out.println(myShortTest2);

OUTPUT

13800
13800

Answer 1

Your first version reads: add up my variables, treat the result as a long, multiply by 10, add 50000 and treat that as a long.

Your second version reads: add up my variables (result is an int), multiply by 10 (which is still an int but might be overflown), add 50000 (still a possibly overflown int) and treat that as a long.

So your fist version starts to treat the sum as a long value and reserves sufficient memory while your second version does this step at the very end, working with lower memory until then.

Answer 2

Whenever an overflow happens, an int will move to the other end of the boundary as seen in the output of the following program:

public class Main {
    public static void main(String[] args) {
        System.out.println(Integer.MAX_VALUE);
        System.out.println(Integer.MAX_VALUE + 1);
        System.out.println(Integer.MIN_VALUE);
        System.out.println(Integer.MIN_VALUE - 1);
    }
}

Output:

2147483647
-2147483648
-2147483648
2147483647

• In the case of test1 , because of casting to long , the result of the intermediate calculation [ 10*(long)(myByte + myShort + myInt) ] was stored as long which can accommodate the result without an overflow and hence you got the correct value.
• In the case of test2 , in lack of proper cast, the result of the intermediate calculation [ 10*(myByte + myShort + myInt) ] was stored as int but the value overflew for int and hence you got the negative value.