Why does int i = 1024 * 1024 * 1024 * 1024 compile without error?
int, java
Solution
There's nothing wrong with that statement; you're just multiplying 4 numbers and assigning it to an int, there just happens to be an overflow. This is different than assigning a single literal, which would be bounds-checked at compile-time.
It is the out-of-bounds literal that causes the error, not the assignment:
System.out.println(2147483648); // error
System.out.println(2147483647 + 1); // no error
By contrast a `long` literal would compile fine:
System.out.println(2147483648L); // no error
Note that, in fact, the result is still computed at compile-time because `1024 * 1024 * 1024 * 1024` is a constant expression:
int i = 1024 * 1024 * 1024 * 1024;
becomes:
0: iconst_0
1: istore_1
Notice that the result (`0`) is simply loaded and stored, and no multiplication takes place.
From JLS §3.10.1 (thanks to @ChrisK for bringing it up in the comments):
It is a compile-time error if a decimal literal of type `int` is larger than `2147483648` (231), or if the decimal literal `2147483648` appears anywhere other than as the operand of the unary minus operator (§15.15.4).
Problem
The limit of `int` is from -2147483648 to 2147483647. If I input ``` int i = 2147483648; ``` then Eclipse will prompt a red underline under "2147483648". But if I do this: ``` int i = 1024 * 1024 * 1024 * 1024; ``` it will compile fine. ``` public class Test { public static void main(String[] args) { int i = 2147483648; // error int j = 1024 * 1024 * 1024 * 1024; // no error } } ``` Maybe it's a basic question in Java, but I have no idea why the second variant produces no error.