CODEWITHSUNDEEP
javabiginteger
user6068334
user6068334

Reputation:

BigInteger is small for my number

I have working with really big number in java. I'm using BigInteger. My number in hex is 0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff

It is about this in decimal: 1.1579208921035625e+77

The code:

public class BigNumber
{
    public static void main(String[] args)
    {
        BigInteger fp = new BigInteger(0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff);
    }
}

And I get error: 

BigNumber.java:5: error: integer number too large: ffffffff00000001000000000000000000000000ffffffffffffffffffffffff
        BigInteger fp = new BigInteger(0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff);

How can I use this big number without loosing the data? Is this possible ? I thought that BigInteger is only limited by memory.

Upvotes: 3

Views: 584

Answers (3)

Sleiman Jneidi
Sleiman Jneidi

Reputation: 23349

The literal 0xfff.. is an integer literal. The compiler detects overflows in literals and this results in a compile error.

You can put it in a string and specify the radix to BigDecimal 

BigInteger fp = new BigInteger("ffffffff00000001000000000000000000000000ffffffffffffffffffffffff",16);

And btw, a constructor that takes an integer as a parameter doesn't exist, but the compiler chose to return this error message.

Upvotes: 2

fabian
fabian

Reputation: 82511

Since you don't use a String to contain the data, the java compiler considers it an int and handles it accordingly. It's obviously too large for the type int:

Use 

new BigInteger("ffffffff00000001000000000000000000000000ffffffffffffffffffffffff", 16)

instead.

Upvotes: 4

Tunaki
Tunaki

Reputation: 137309

That's because you need to use the String constructor, that also takes a radix. In this case, the radix is 16.

public static void main(String[] args) {
    BigInteger fp = new BigInteger("ffffffff00000001000000000000000000000000ffffffffffffffffffffffff", 16);
    System.out.println(fp);
}

In your snippet, the hexadecimal value is converted into an int because it is an integer literal and since it is too big to fit in an int, an error is raised.

Upvotes: 10

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