Is there a difference on how java performs operations using shortcut operators from the regular ones?

Question

I am working on a java program concerning the pascal's triangle.

So this is how it is coded:

for(int i = 0; i < 5; i++){
    for(int j = 0, x = 1; j <= i; j++){
        System.out.print(x + " ");
        x = x * (i - j) / (j + 1);
    }
    System.out.println();
}

and it shows:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

But when I tried to change the code to:

for(int i = 0; i < 5; i++){
    for(int j = 0, x = 1; j <= i; j++){
        System.out.print(x + " ");
        x *= (i - j) / (j + 1);
    }
    System.out.println();
}

and as you may have noticed, only the operator has changed to *=, but the result is:

1
1 1
1 2 0
1 3 3 0
1 4 4 0 0

Any idea what must have happened? Thanks in advance!

Answer 1

Looks like integer division versus order of operations. Try adding some parenthesis and I think you will eventually achieve the same results. If you, say, divide 2/3 in integers, you get 0. So it matters if you do some multiplying first.

Answer 2

It's because you're using integer arithmetic in the wrong order.

x *= (i - j) / (j + 1);

is the same as

x = x * ((i - j) / (j + 1));

The brackets are important. (i - j) / (j + 1) is in most cases not a whole number, but java rounds it to an integer anyway.

The way you did it first

x = x * (i - j) / (j + 1);

the multiplication happens before the division, so you don't get any rounding errors.

Answer 3

You switched the high precedence * for a low precedence *= resulting in

x = x * ((i - j) / (j + 1));

in stead of

x = (x * (i - j)) / (j + 1);

which you probably wanted.