Having trouble understanding Java Overflow and Compile error

Question

I am a Java newbie and studying int overflow! While playing with some integers I was surprised with some weird results

    int x = 2147483647 + 1;    
    x == > - 2147483648 

    int x = 2147483647 + 2;
    x ==> -2147483647
    
    int x = 2147483647 + 2147483647;
    x ==> -2

    int x = 2147483647 + 2147483648; 
    **compile error**

I thought integer overflow would not cause any compile error. Also, it is hard for me to understand how outputs for the overflow are calculated (ex. why int x = 2147483647 + 1 // x ==> -2147483648) Can anybody please expalin the logic of these results?

Thanks!!

Answer 1

From the language spec:

The largest decimal literal of type int is 2147483648 (2^31).

All decimal literals from 0 to 2147483647 may appear anywhere an int literal may appear. The decimal literal 2147483648 may appear only as the operand of the unary minus operator - (§15.15.4).

It is a compile-time error if the decimal literal 2147483648 appears anywhere other than as the operand of the unary minus operator; or if a decimal literal of type int is larger than 2147483648 (2^31).

You can't use 2147483648 as an int literal because an int literal is an int expression, and thus must have an int value; but 2147483648 is too large to be used as an int value.