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javaphpfloor

Trying to understand this code behaviour

Today i was working on a PHP project and came across this code behaviour

<?php
$x = 5.5;
$y = 0;
echo $z = floor($x * $y * -1);
?>

This gave the output of -0 . Can anyone shed light on why this echoes -0 . But I expected 0 Only when adding floor this appears to happen. I tried the same in java.

class Sample {
    public static void main(String[] args) {
        float x =5.5f;
        int y = 0;
        System.out.println(Math.floor(x*y*-1));
    }
}

This also prints -0.0 .

Solution

  • float and double have both a positive 0 and a negative 0. When you multiple 0 * -1 you get -0 as specified in the IEEE 754 standard.

    Note: 1/0 is positive infinity but 1/-0 is negative infinity.

    You can see that http://ideone.com/tBd41l

    System.out.println(0f * -1);

    prints

    -0.0

    Math.floor is not required.