CODEWITHSUNDEEP
java
Bathsheba
Bathsheba

Reputation: 234725

Behaviour when incrementing a Java long past the maximum

Consider:

long n = Long.MAX_VALUE;
++n;

With my JVM, this wraps round to Long.MIN_VALUE. Is this the defined behaviour? Will some JVMs throw an exception instead?

Upvotes: 0

Views: 116

Answers (4)

Narendra Pathai
Narendra Pathai

Reputation: 41955

It is expected : integer types overflow and underflow silently

JLS: Primitive Types and Values quotes that

The integer operators do not indicate overflow or underflow in any way.

So as the arithmetic types in Java are signed, adding 1 to Long.MAX_VALUE causes overflow and it goes to Long.MIN_VALUE as Java uses 2's complement

As the JLS quotes all the integer operators such as ++ will silently overflow the value and not cause any exception

Upvotes: 1

Olimpiu POP
Olimpiu POP

Reputation: 5067

As pointed out in the JLS :

The integer operators do not indicate overflow or underflow in any way.

An integer operator can throw an exception (§11) for the following reasons:

Any integer operator can throw a NullPointerException if unboxing conversion (§5.1.8) of a null reference is required.

The integer divide operator / (§15.17.2) and the integer remainder operator % (§15.17.3) can throw an ArithmeticException if the right-hand operand is zero.

The increment and decrement operators ++ (§15.14.2, §15.15.1) and -- (§15.14.3, §15.15.2) can throw an OutOfMemoryError if boxing conversion (§5.1.7) is required and there is not sufficient memory available to perform the conversion.

and addition(as pointed out by Svetelin Zarev):

If an integer addition overflows, then the result is the low-order bits of the mathematical sum as represented in some sufficiently large two's-complement format. If overflow occurs, then the sign of the result is not the same as the sign of the mathematical sum of the two operand values.

I think it is safe to assume that no exception will be thrown in this case and everytime it will overflow to Long.MIN_VALUE.

Upvotes: 0

kai
kai

Reputation: 6887

Long.MAX_VALUE 9223372036854775807 = +(2^63-1) : 

01111111.11111111.11111111.11111111.11111111.11111111.11111111.11111111

++n will increment this binary number and so the sign bit will be used:

Long.MIN_VALUE -9223372036854775808 = -(2^63) : 

10000000.00000000.00000000.00000000.00000000.00000000.00000000.00000000

thats not the behaviour of the jvm, thats the behaviour of an icrementation of an binary number.

Upvotes: 2

Svetlin Zarev
Svetlin Zarev

Reputation: 15683

From the JLS:

If an integer addition overflows, then the result is the low-order bits of the mathematical sum as represented in some sufficiently large two's-complement format. If overflow occurs, then the sign of the result is not the same as the sign of the mathematical sum of the two operand values.

Source: http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.18.2

Upvotes: 3

Related Questions