Java - long displays wrong values

Question

I have to create a program that goes from 1 to 100 and calculates the separate product of the numbers that have from the division with 4 the rest 0, 1, 2, 3, so basically i have to calculate 4 different products.

I tried solving this problem in two ways both ways display the wrong results:

First attempt:

    public class prodmod4 {
public static void main(String[] args){
    int i, k=0;
    long P=1;
    while(k<4){
        for(i=1;i<=100;i++){
            if(i%4==k){
            P=P*i;
            }
        }
        System.out.printf("Produsul numerelor care au restul %d este: %d\n", k, P);
        i=1;
        P=1;
        k++;
    }
}
}

When i run this program it gives me:

The product of numbers with rest 0 is 0, rest 1 -6661643765781265135, rest 2 -2885954222765899776, rest 3 -9150527387120197261

Second attempt:

public class prodmod4v2 {
public static void main(String[] args){
    int i;
    long zero=1, unu=1, doi=1, trei=1;
    for(i=1;i<=100;i++){
        switch(i%4){
        case 0:
            zero=zero*i;
            break;
        case 1:
            unu=unu*i;
            break;
        case 2:
            doi=doi*i;
            break;
        case 3:
            trei=trei*i;
            break;
        default:
            break;
        }
    }
    System.out.printf("produsul numerelor care au resturile 0,1,2,3 sunt:\n zero:%d\n unu:%d\n doi:%d\n trei:%d\n", zero, unu, doi, trei);
}
}

When i run this i have the same output as in the first attempt.

Thank you in advance!

This could be due to simple overflow of the longs. Try printing your statement at the end of each loop iteration. — River, Jul 14 '15 at 20:51
Check this SO question and its answers: http://stackoverflow.com/questions/3001836/how-does-java-handle-integer-underflows-and-overflows-and-how-would-you-check-fo — Chthonic Project, Jul 14 '15 at 20:53

score 5 · Accepted Answer · edited May 23 '17 at 11:58

5

Your numbers are too big for a mere long, which supports numbers up to 2^63 - 1. When this happens, the result is said to overflow - which means that the results are not what you'd expect them to be. To solve this, you must use something that supports larger numbers, like BigInteger or BigDecimal.

edited May 23 '17 at 11:58

Community

1
1

answered Jul 14 '15 at 20:48

Glorfindel

21,988
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[BigInteger](http://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html) is what is wanted here, since the numbers involved are products of integers, and thus will always be integral. – David Conrad Jul 14 '15 at 22:36
I declared them as BitInteger and now it trows me the following error: "the method multiply(long) from the type BitInteger is not visible" at the lines where i have to multiply. Thanks is advance! – Nerozx Jul 15 '15 at 13:06
You can only multiply BigIntegers like this: `new BigInteger(4).multiply(new BigInteger(5));` – Glorfindel Jul 15 '15 at 13:07
The way i did it was: zero=zero.multiply(i); and it gives me that error. So i can't multiply BitIntegers with integers? – Nerozx Jul 15 '15 at 13:13
zero = zero.multiply(new BigInteger(i)) – Glorfindel Jul 15 '15 at 13:14
Or bigint.multiply(BigInteger.valueOf(i)) – David Conrad Jul 15 '15 at 13:29
@DavidConrad it worked with your solution! Thanks for all your help, everyone! – Nerozx Jul 15 '15 at 14:01

Java - long displays wrong values

1 Answers1