PHP in_array does not seem to work

Question

i have two files: index.php and logspace.php. In logspace.php i have a function logspace that returns an array. I call this function from index.php. Then I want to check if this returned array has certain value but 'in_array' function doesn't work.

Code in index.php:

<?php
include 'logSpace.php';
$tmp=logspace(0.1,10,9);
$Tr=0.1;

if(in_array($Tr,$tmp))
{
    echo 'true';
}
else
{
    echo 'false';
}
?>

I always get 'false' even though value is clearly in the array:

var_dumb($Tr);

float(0.1)

var_dump($tmp);

array(10) { [0]=> float(0.1) [1]=> float(0.16681005372001) [2]=> float(0.27825594022071) [3]=> float(0.46415888336128) [4]=> float(0.77426368268113) [5]=> float(1.2915496650149) [6]=> float(2.1544346900319) [7]=> float(3.5938136638046) [8]=> float(5.9948425031894) [9]=> float(10) }

Code in logSpace.php

function logspace($start,$end,$num)
{
    $arr=array();
    $logMin=log($start);
    $logMax=log($end);
    $delta=($logMax-$logMin)/$num;
    $accDelta=0;

    for($i=0;$i<=$num;$i++)
    {
        $num_i=pow(M_E,$logMin+$accDelta);
        $arr[]=$num_i;
        $accDelta+=$delta;
    }        
return $arr;
}
?>

@SureshKamrushi: doesn't matter. only PHP variables are case sensitive. functions/language syntax isn't. — Marc B, Nov 12 '13 at 14:53
This is most likely caused by the floating point, have you tryed storing the numbers as strings? (example: $arr = array("0.1"); echo in_array("0.1",$arr) ? "true" : "false";) — cernunnos, Nov 12 '13 at 14:54
what's going on in lines 9 and 13 of logspace.php? they don't seem valid — msturdy, Nov 12 '13 at 14:55
`$accDelta+=delta;` should be `$accDelta+=$delta;` and in `$accDelta=$0;` `$0` doesn't exist. i guess it should be `$accDelta=0;` — Ed T., Nov 12 '13 at 14:57

score 4 · Accepted Answer · answered Nov 12 '13 at 15:04

in_array is going to be testing equality at some point, but your values are floating-point numbers. Equality between floats in any computer language is touchy at best, because of limited precision.

1/10 in base-2 is classically compared to 1/3 in base-10. You can't write 1/3 as a decimal using finite precision and be exact (0.3333...), and you have the same problem with 0.1 on a computer. The solution is either avoiding floats entirely (storing values as strings, storing integer multiples of your floats, calculating your float values on-the-fly, etc.), or sticking strictly to inequalities, rather than equality operations.

For example:

$epsilon = 0.00001;
$target_num = 0.1;

foreach ($tmp as $flt)
{
    if ($flt - $epsilon < $target_num && $target_num < $flt + $epsilon)
    {
        return true; // ~0.1 is in $tmp
    }
}
return false; // ~0.1 is not in $tmp

This is it, thanks. I converted each element of an array to string with $tmp=array_map('strval',$tmp); before calling in_array. — micco, Nov 12 '13 at 15:14

PHP in_array does not seem to work

1 Answers1

Linked

Related