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javatypesintlong-integer
Pubudu Dodangoda
Pubudu Dodangoda

Reputation: 2884

java cannot handle a 32 bit number

I am trying to assign 4294967295 to a long. That is (2^32-1) java(netbeans) gives the following error message "integer number too large"

in fact I tried to figure out the largest number that an int can handle(did it manually by hand) and found that it is 2147483647 (of course as obvious it is 2^31-1)

But surprisingly I found out that even the long type cannot handle a number larger than that. Isn't there any difference between int and long?java doc says long is 64 bit

Am I missing something?

Upvotes: 2

Views: 2367

Answers (3)

Pascal
Pascal

Reputation: 380

From The Java™ Tutorials - Primitive Data Types:

int: By default, the int data type is a 32-bit signed two's complement integer, which has a minimum value of -231 and a maximum value of 231-1. In Java SE 8 and later, you can use the int data type to represent an unsigned 32-bit integer, which has a minimum value of 0 and a maximum value of 232-1. Use the Integer class to use int data type as an unsigned integer. See the section The Number Classes for more information. Static methods like compareUnsigned, divideUnsigned etc have been added to the Integer class to support the arithmetic operations for unsigned integers.

So 232-1 = 4294967295

Upvotes: 0

Jens
Jens

Reputation: 69495

Use the lower l to show the compiler that it is an long value.

long l = 4294967295l ;

Upvotes: 3

Jon Skeet
Jon Skeet

Reputation: 1503579

The problem is that you're using 4294967295 as an int literal - but it's not a valid int value. You want it to be a long literal, so you need to put the L suffix on it. This is fine:

long x = 4294967295L;

From JLS section 3.10.1:

An integer literal is of type long if it is suffixed with an ASCII letter L or l (ell); otherwise it is of type int (§4.2.1).

Upvotes: 12

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