Why does my variable resut in an overflow in Java?

Question

Here is my code num1 etc are int variable

double numProduct1;
 numProduct1 = (num1 * num2 * num3 * num4);

System.out.printf("%.3f ", numProduct1);

Inputs are respectively (num1,num2,num3,num4): 100000 200000 300000 500000

my code would output -1679818752.000 instead of 3000000000000000000000.000

Because 3000000000000000000000 is out of the range for an `int`. — Sweeper, Sep 27 '20 at 05:03
How Do I fix that? I thought having a double as a variable would work fine? — BreakTheGreat, Sep 27 '20 at 05:06

Sweeper · Accepted Answer · 2020-09-27T05:14:58.087

You thought that assigning the product to a double will allow you to use the whole range of double, but that isn't the case. You are only converting the result to double. The multiplication is still carried out with int, and the int range applies.

Essentially, this is the multiplication version of this common question (well, not exactly :D).

The fix is similar, simply make the first operand double (e.g. by casting):

numProduct1 = ((double)num1 * num2 * num3 * num4);

The expression is now a double multiplied by an int, and numeric promotion occurs, promoting the other numbers to double as well.

Why does my variable resut in an overflow in Java?

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