PHP round to integer

Question

I want to round a number and I need a proper integer because I want to use it as an array key. The first "solution" that comes to mind is:

$key = (int)round($number)

However, I am unsure if this will always work. As far as I know (int) just truncates any decimals and since round($number) returns a float with theoretically limited precision, is it possible that round($number) returns something like 7.999999... and then $key is 7 instead of 8?

If this problem actually exists (I don't know how to test for it), how can it be solved? Maybe:

$key = (int)(round($number) + 0.0000000000000000001) // number of zeros chosen arbitrarily

Is there a better solution than this?

Integers stored within floats are always accurate, up to around 2^51, which is much more than can be stored in an int anyway. You are worrying over nothing. — Niet the Dark Absol
– Niet the Dark Absol, Commented Aug 3, 2016 at 14:22
@NiettheDarkAbsol Oops, I actually knew that (from Javascript) but didn't make the connection. You should make that an answer. BTW, in JS it's 2^53-1 iirc. — AndreKR
– AndreKR, Commented Aug 3, 2016 at 14:23
When you use rounded numbers as key, you run in trouble if you have more then one 7.x numbers with the rounded result of 8. — u-nik
– u-nik, Commented Aug 3, 2016 at 14:25
@u-nik It's exactly my intention that 7.4 and 7.6 end up with the same key. :) — AndreKR
– AndreKR, Commented Aug 3, 2016 at 14:27
There are three functions to handle rounding numbers: round, [ceil][] (round up), [floor][] (round down). I'm unsure which one you want to use. [ceil]: secure.php.net/ceil [floor]: secure.php.net/floor — Exagone313
– Exagone313, Commented Aug 3, 2016 at 14:28

Community · Accepted Answer · 2017-05-23 12:18:26Z

63

To round floats properly, you can use:

ceil($number): round up
round($number, 0): round to the nearest integer
floor($number): round down

Those functions return float, but from Niet the Dark Absol comment: "Integers stored within floats are always accurate, up to around 2^51, which is much more than can be stored in an int anyway."

edited May 23, 2017 at 12:18

CommunityBot

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answered Aug 3, 2016 at 14:26

Mistalis

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1 Comment

Donald Duck is with Ukraine Over a year ago

The problem with these is that their return type is float, not int.

Dharman · Accepted Answer · 2025-01-29 20:52:40Z

9

round(), without a precision set always rounds to the nearest whole number. By default, round rounds to zero decimal places.

So:

$int = 8.998988776636;
round($int) //Will always be 9

$int = 8.344473773737377474;
round($int) //will always be 8

So, if your goal is to use this as a key for an array, this should be fine.

You can, of course, use modes and precision to specify exactly how you want round() to behave. See this.

You might actually be more interested in intval:

echo intval(round(4.7)); //returns int 5
echo intval(round(4.3)); // returns int 4

edited Jan 29 at 20:52

Dharman♦

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answered Aug 3, 2016 at 14:28

Zac Brown

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3 Comments

AndreKR Over a year ago

Nope, intval(4.7) == 4, which is not what I want.

Zac Brown Over a year ago

Yea, with intval by itself. Combine it with round. I've updated my answer

AndreKR Over a year ago

That suffers from the same (non-)problem that I described.

Chris Seufert · Accepted Answer · 2019-03-18 22:43:33Z

6

What about simply adding 1/2 before casting to an int?

eg:

$int = (int) ($float + 0.5);

This should give a predictable result.

answered Mar 18, 2019 at 22:43

Chris Seufert

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1 Comment

Charliexyx Over a year ago

why only three upvotes? I mean, yes, 2^51 as described in accepted answer is a lot, but still this technique is even more correct, right?

Chandan Sharma · Accepted Answer · 2020-09-22 12:50:40Z

2

For My Case, I have to make whole number by float or decimal type number. By these way i solved my problem. Hope It works For You.

$value1 = "46.2";
$value2 = "46.8";


// If we print by round()
echo round( $value1 );    //return float 46.0
echo round( $value2 );    //return float 47.0


// To Get the integer value
echo intval(round( $value1 ));   // return int 46
echo intval(round( $value2 ));   // return int 47

answered Sep 22, 2020 at 12:50

Chandan Sharma

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1 Comment

Chandan Sharma Over a year ago

Please comment if have any Issue with the code. will try to update for better understand.

AndreKR · Accepted Answer · 2016-08-03 14:40:54Z

1

Integers stored within floats are always accurate, up to around 2⁵³, which is much more than can be stored in an int anyway. I am worrying over nothing.

answered Aug 3, 2016 at 14:40

AndreKR

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Comments

JSowa · Accepted Answer · 2024-10-03 10:27:33Z

1

Update PHP 8.4

PHP 8.4 introduces new 3 functions for arbitrary precision rounding in extension BCMath: bcceil(), bcround(), bcfloor().

bcround('0.285', 2, \RoundingMode::HalfAwayFromZero); // string(4) "0.29"

bcfloor('13.0915'); // string(2) "13"

bcceil('24.00001'); // string(2) "25"

If you set up the precision to 0, you can easily cast the string to an integer without a floating-point error.

More information:

PHP RFC: Adding bcround, bcfloor and bcceil to BCMath

Moreover, you can make use of the new 4 rounding modes added to the standard library which can be used with the round() function and rounding functions from BCmath.

round(5.5, mode: \RoundingMode::HalfAwayFromZero); // 6
round(5.5, mode: \RoundingMode::HalfTowardsZero); // 5
round(5.5, mode: \RoundingMode::HalfEven); // 6
round(5.5, mode: \RoundingMode::HalfOdd); // 6

//introduced in PHP 8.4
round(5.5, mode: \RoundingMode::TowardsZero); // 5
round(5.5, mode: \RoundingMode::AwayFromZero); // 6
round(5.5, mode: \RoundingMode::NegativeInfinity); // 5
round(5.5, mode: \RoundingMode::PositiveInfinity); // 6

More information:

edited Oct 3, 2024 at 10:27

answered Oct 2, 2024 at 16:42

JSowa

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2 Comments

AndreKR Over a year ago

They all still return a float.

JSowa Over a year ago

You are right. I forgot to put all changes.

Евгений Криворучко · Accepted Answer · 2021-08-26 12:01:57Z

-1

My solution:

function money_round(float $val, int $precision = 0): float|int
{
    $pow = pow(10, $precision);

    $result = (float)(intval((string)($val * $pow)) / $pow);
    if (str_contains((string)$result, '.')) {
        return (float)(intval((string)($val * $pow)) / $pow);
    }
    else {
        return (int)(intval((string)($val * $pow)) / $pow);
    }
}

answered Aug 26, 2021 at 12:01

Евгений Криворучко

1

1 Comment

Roman Over a year ago

Welcome to StackOverflow! Please provide an explanation for your code.

Cyril Joudieh · Accepted Answer · 2016-08-03 14:25:31Z

-3

Round to the nearest integer

$key = round($number, 0);

answered Aug 3, 2016 at 14:25

Cyril Joudieh

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3 Comments

AndreKR Over a year ago

It returns a float.

Cyril Joudieh Over a year ago

How about $key = intval(round($number, 0)) or $key = (int)round($number, 0)

AndreKR Over a year ago

Did you even read my question? I specifically gave (int)round($number) as an example of what I thought is not enough.

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