Incorrect Multiplication

Question

I am trying to solve this one particular problem involving basic math. My algorithm works for every other test cases and thus i know its correct. But for one particular input in one of the test cases, the result is wrong. I checked the values by printing them and values are:

a = 1000000000 and b = 999999999

When I do

System.out.print(a*b);

it returns 1808348672.

I have tried doing this:

long ans = (long)a*b;
System.out.print(ans);

It still returns 1808348672

That’s a bad duplicate — it doesn’t answer this question. The real issue is that the problem isn’t reproducible because OP’s code works. — Konrad Rudolph, Sep 07 '20 at 07:35
@Mureinik Please leave it alone. The whole problem here is that there was *no* relevant [tag:casting]. Just a non-reproducible problem. — user207421, Sep 08 '20 at 10:28

score 5 · Answer 1 · answered Sep 07 '20 at 07:42

5

I have tried doing this:
long ans = (long)a*b;
System.out.print(ans);
It still returns 1808348672

No, it doesn’t:

class Foo {
    public static void main(String[] args) {
        int a = 1000000000;
        int b = 999999999;
        long ans = (long)a*b;
        System.out.print(ans);
    }
}

javac foo.java && java Foo
999999999000000000

What you probably tried instead is

long ans = (long) (a*b);

This will first perform integer multiplication, and then cast the result to long. That doesn’t work since the integer multiplication overflows before your cast happens.

answered Sep 07 '20 at 07:42

Konrad Rudolph

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My bad. I used int type for ans and tried to store long in an int variable. – Jaideep Kular Sep 07 '20 at 07:47
@JaideepKular That would not have compiled without yet another cast. – user207421 Sep 07 '20 at 07:48

score 2 · Answer 2 · answered Sep 07 '20 at 07:29

2

a and b are ints, so their multiplication will also be an int, which overflows before being promoted to a long by the explicit cast.

You can define them as longs or cast them individually to longs before performing the multiplication.

answered Sep 07 '20 at 07:29

Mureinik

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It worked by casting each of them individually. But why do I need to cast them individually? I tried to cast the whole operation and it didn't work. – Jaideep Kular Sep 07 '20 at 07:40
1

@JaideepKular casting the result is too late - multiplying two `int` will result in an `int`, which already overflows before you cast it. If you cast them individually, you'd be multiplying two `long`s, which result in a `long` that can accommodate the result without overflowing. – Mureinik Sep 07 '20 at 07:53

Incorrect Multiplication

2 Answers2