Overflow while bit shifting in java

Question

I have situation which I dont undestand. When I do

 System.out.println(1<<30);

I got very big positive number 1073741824. But when I do

 System.out.println(1<<31);

I got very low number -2147483648. As I read long type is signed 64 bit long. Max number is 2^63 -1. So here I dont undestand why my shift goes negative(((( Doing

System.out.println((long)(1<<31));

also gives the same result(((

Yep, you are right. But why System.out.println((long)(1<<31)); does not work either??? — ovod
– ovod, Commented Dec 10, 2014 at 14:39
because you're casting too late, when it's already MIN_VALUE. See the answer, it's right. — Denys Séguret
– Denys Séguret, Commented Dec 10, 2014 at 14:39
Because you cast the result (an int) not the operands. The expression (1<<31) is already an integer. — PeterMmm
– PeterMmm, Commented Dec 10, 2014 at 14:40
Thank you, very much. Very interesting situation for me. I created 100 tests for my little program and 1 of 100 didnt pass. Now I see why. Spent a day to catch it))) — ovod
– ovod, Commented Dec 10, 2014 at 14:43

fge · Accepted Answer · 2014-12-10 14:49:44Z

6

1 << 31 is an int, not a long. And what you have is not really an overflow, but Integer.MIN_VALUE (sign bit set, all the rest zeroes).

Unless you suffix them appropriately, any numeric constant in Java is an int by default. If you want a long, you have to tell the compiler that you want a long: 1L << 31 (note the L).

And note that even (long) (1 << 31) wouldn't work; even in this scenario, 1 << 31 remains an int. Casting "after the fact" doesn't help ;)

edited Dec 10, 2014 at 14:49

answered Dec 10, 2014 at 14:36

fge

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Evgeniy Dorofeev · Accepted Answer · 2014-12-10 14:43:50Z

0

When we shift 1 by 31 to the left we get 0x80000000, this is Integer.MIN_VALUE = -2147483648. See Integer class.

answered Dec 10, 2014 at 14:43

Evgeniy Dorofeev

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