Decimal positions in an array? (Java)

Question

I've seen this code and it's supposed to be fine, I don't understand why if for example "left + right" is an odd number, doing "a[middle] == 0" doesn't give error because it doesn't exist in arrays decimal positions. Thank you.

/** Searches the value 0 within a[left..right].
* Precondition: 0 <= left, right < a.length,
* a[left..right] is sorted in ascending order. */

public static int intercepts(double[] a, int left, int right) {
if (left > right) { return -1; }
else {
int middle = (left + right) / 2;
if (a[middle] == 0) { return middle; }
else if (a[middle] > 0) {
return intercepts(a, left, middle - 1);
} else {
return intercepts(a, middle + 1, right); }
}
}

In Java, applying `/` to two integers gives an integer result. — Dawood ibn Kareem, Mar 31 '19 at 22:50

Marvin · Accepted Answer · 2019-03-31T23:07:19.247

3

left, right and 2 are all integers, therefore an integer division will be performed and the result will also be an integer. Integer division rounds toward 0, so all decimals are simply cut off.

See for yourself:

public static void main (String[] args) {
    int middle = 5 / 2;
    System.out.println(middle); // 2
}

edited Mar 31 '19 at 23:07

answered Mar 31 '19 at 22:50

Marvin

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score 3 · Answer 2 · answered Mar 31 '19 at 23:08

Since the two operands (left and right) are integers then integer arithmetic is used. When the division is applied, the remainder of the result will be "rounded" down to zero.

Example

5/2 = 2
7/3 = 2 
8/3 = 2

See the following article for a better explanation. Int division: Why is the result of 1/3 == 0?

Decimal positions in an array? (Java)

2 Answers2