Dividing by zero without ArithmeticException

Question

When trying to write a truthtable I tried to figure out the algorithm, and I found one that works. However now I don't understand why it doesn't throw a ArithmeticException.

for(int i=0; i<Math.pow(2, numberOfInputs()); i++)
    {
        for(int j=0; j<numberOfInputs(); j++)
        {
            System.out.print("" + i/(int) Math.pow(2, j)%2 + "\t");

        }
        System.out.println("");
    }

When trying this with numberOfInputs() returning 2, I get:

Truthtable
0   0   
1   0   
0   1   
1   1

But I dont't understand how this can be since Math.pow(2, j)%2 will be =0 multiple times?!

Look up "Java operator precedence". – user202729 Mar 10 '19 at 11:16 — user202729, Mar 10 '19 at 11:16
The code is equivalent to `(i/2^j) % 2` – Minn Mar 10 '19 at 11:23 — Minn, Mar 10 '19 at 11:23
stop using `Math.pow(2, j)`. Use `1 << j` instead – phuclv Mar 14 '20 at 05:58 — phuclv, Mar 14 '20 at 05:58

score 2 · Answer 1 · answered Mar 10 '19 at 11:20

2

It's because of the precedence, this is executed first: i/(int) Math.pow(2, j), then modulo. The first expression is never a division by zero. More about it in the documentation.

answered Mar 10 '19 at 11:20

Andronicus

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Dividing by zero without ArithmeticException

1 Answers1