PHP code returns value in float

Question

In the following code, why does $final returns value as float?

When the calculated value is 6, it should return as integer, otherwise float.

How can I do that?

$x = 62;
$round = 5 * round($x / 5);
$final = $round/10;     
var_dump($final); 
float(6)

Edit: Sorry if my question is not clear. I need to find out if the $final has any decimal value or not. So in order to find out that, I was using is_float function, but that always returns true because above variable returns the value in float always. Hope my question is a bit more clear.

php types the variables along with their usage. There's no strong beforehand typing. You could use intval afterwards though. (if intval($final) == $final)... — Sebas
– Sebas, Commented Sep 30, 2013 at 11:30
So you have to check whether you got exactly 6 (and intval($final) that), otherwise got float ... where is the problem? — djot
– djot, Commented Sep 30, 2013 at 11:31
If you are working with float numbers you should consider using the BC Math functions - php.net/manual/en/book.bc.php — hdvianna
– hdvianna, Commented Sep 30, 2013 at 11:51

hek2mgl · Accepted Answer · 2013-09-30 11:39:34Z

2

The return type of round is float, as the manual states. You need to cast the result to integer:

$result = (integer) round($value);

Update:

Although I think a method should return either floats or integers and not change this depending on the result's value, you could try something like this:

if((integer) $result == $result) {
    $result = (integer) $result;
}
return $result;

edited Sep 30, 2013 at 11:39

answered Sep 30, 2013 at 11:31

hek2mgl

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8 Comments

djot Over a year ago

This does not return float (as wanted by OP), if value is e.g. 6.21, it should(!) return as float.

hek2mgl Over a year ago

how should the result of round being 6.21 ?

djot Over a year ago

I understood the question like if the result is 6.00 return INT 6, if the result is 6.21, return float 6.21.

hek2mgl Over a year ago

If you ask me, the question is questionable ;)

djot Over a year ago

This is clear to me, at least me ;) ... "When the calculated value is 6, it should return as integer, otherwise float."

|

djot · Accepted Answer · 2013-09-30 11:52:06Z

1

See this code live here:

http://sandbox.onlinephpfunctions.com/code/f8f383604cf848ab63534da69295f8482528e4ce

-

<?php

    $final = function($num) {

      $calc = (5 * ($num / 5)) / 10;

      if (intval($calc) == $calc) { settype($calc, "integer"); }
      else { settype($calc, "float"); } // might not need that line
      return $calc;
    };

    var_dump($final(62)); print '<br />';
    var_dump($final(60)); print '<br />';
    var_dump($final(59)); print '<br />';

?>

Besides, in the OPs code, this line (the round())is incorrect:

$round = 5 * round($x / 5);

edited Sep 30, 2013 at 11:52

answered Sep 30, 2013 at 11:45

djot

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1 Comment

djot Over a year ago

Thanks to myself for spending (did I say waisting?) my time on that ;)

Daryl Gill · Accepted Answer · 2013-09-30 11:46:41Z

0

Little bit long, but something like this should do the trick

function ReturnType($FinalVal){
 switch (gettype($FinalVal)){
  case 'integer':
     return (int) $FinalVal;
  case 'double':
    return (float) $FinalVal;

 }
}

 $x = 62;
$round = 5 * round($x / 5);
$final = $round/10;     
$Value = ReturnType($final);

answered Sep 30, 2013 at 11:46

Daryl Gill

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1 Comment

djot Over a year ago

No, guess not correct, see sandbox.onlinephpfunctions.com/code/…

david · Accepted Answer · 2013-09-30 11:32:50Z

-1

You must casting like this

$round = 5 * (int)(round($x / 5));

round($x / 5) returns float. If you don't cast it to int then the result would be float

answered Sep 30, 2013 at 11:32

david

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1 Comment

djot Over a year ago

and this was NOT the question ... it should return int or float, not only int ... 5* INT is ... still int

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