CODEWITHSUNDEEP
javainteger
user446654
user446654

Reputation: 1141

"Integer number too large" error message for 600851475143

public class Three {
    public static void main(String[] args) {
        Three obj = new Three();
        obj.function(600851475143);
    }

    private Long function(long  i) {
        Stack<Long> stack = new Stack<Long>();

        for (long j = 2; j <= i; j++) {
            if (i % j == 0) {
                stack.push(j);
            }
        }
        return stack.pop();
    }
}

When the code above is run, it produces an error on the line obj.function(600851475143);. Why?

Upvotes: 111

Views: 221352

Answers (8)

Milen.Jeremic
Milen.Jeremic

Reputation: 31

Or, you can declare input number as long, and then let it do the code tango :D ...

public static void main(String[] args) {

    Scanner in = new Scanner(System.in);
    System.out.println("Enter a number");
    long n = in.nextLong();

    for (long i = 2; i <= n; i++) {
        while (n % i == 0) {
            System.out.print(", " + i);
            n /= i;
        }
    }
}

Upvotes: 3

Anand Undavia
Anand Undavia

Reputation: 3543

Apart from all the other answers, what you can do is : 

long l = Long.parseLong("600851475143");

for example : 

obj.function(Long.parseLong("600851475143"));

Upvotes: 4

Yuliy
Yuliy

Reputation: 17728

600851475143 cannot be represented as a 32-bit integer (type int). It can be represented as a 64-bit integer (type long). long literals in Java end with an "L": 600851475143L

Upvotes: 251

JVM
JVM

Reputation: 229

At compile time the number "600851475143" is represented in 32-bit integer, try long literal instead at the end of your number to get over from this problem.

Upvotes: 5

Thilo
Thilo

Reputation: 262684

You need to use a long literal: 

obj.function(600851475143l);  // note the "l" at the end

But I would expect that function to run out of memory (or time) ...

Upvotes: 38

josefx
josefx

Reputation: 15656

The java compiler tries to interpret 600851475143 as a constant value of type int by default. This causes an error since 600851475143 can not be represented with an int.

To tell the compiler that you want the number interpretet as a long you have to add either l or L after it. Your number should then look like this 600851475143L.

Since some Fonts make it hard to distinguish "1" and lower case "l" from each other you should always use the upper case "L".

Upvotes: 16

Andre Holzner
Andre Holzner

Reputation: 18695

You need 40 bits to represent the integer literal 600851475143. In Java, the maximum integer value is 2^31-1 however (i.e. integers are 32 bit, see http://download.oracle.com/javase/1.4.2/docs/api/java/lang/Integer.html).

This has nothing to do with function. Try using a long integer literal instead (as suggested in the other answers).

Upvotes: 7

Roman
Roman

Reputation: 66196

Append suffix L: 23423429L.

By default, java interpret all numeral literals as 32-bit integer values. If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.

Upvotes: 91

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