fmod
(PHP 4 >= 4.2.0, PHP 5, PHP 7, PHP 8)
fmod — 返回除法的浮点数余数
说明
返回被除数（num1）除以除数（num2）所得的浮点数余数。余数（r）的定义是：num1
= i * num2 + r，其中 i 是整数。如果 num2
是非零值，则 r 和
num1 的符号相同并且其数量值小于
num2。
参数
num1-
被除数
num2-
除数
返回值
num1/num2 的浮点数余数。
示例
示例 #1 fmod() 的使用
<?php
$x = 5.7;
$y = 1.3;
$r = fmod($x, $y);
// $r equals 0.5, because 4 * 1.3 + 0.5 = 5.7
?>
＋添加备注
用户贡献的备注 7 notes
nospam at neonit dot de ¶
8 years ago
Note that fmod does not behave like a similar function written in PHP itself does due to the lack of fixing floating point representation errors.
Have a look at this:
<?php
var_dump(10 / (10 / 3) === 3.0); // bool(true)
var_dump(fmod(10, 10 / 3)); // float(3.3333333333333)
var_dump(fmod(10, 10 / 3) < 10 / 3); // bool(true)
?>
Internally there is no way of exactly representing the result of 10 / 3, so it will always be a bit above or below the actual result. In this case, the example proves it being a bit above the actual result.
PHP seems quite good at auto-fixing floating point representation errors so they behave like the user would expect it. That's why the first line yields true, although the result is slightly below 3 (like 2.9999999999[something]). I failed to trick PHP into rounding or cropping the result to 2.
However, fmod seems to not apply these fixes during calculations. From 10 / 3 it gets a value slightly below 3, floors it to 2 and returns 10 - 2 * 10 / 3, which is slightly less than the actual result of 10 / 3, but looks like 10 / 3 (third line).
Unfortunately, this is not the expected result. See other notes for high quality fixes.
jphansen at uga dot edu ¶
20 years ago
fmod() does not mirror a calculator's mod function. For example, fmod(.25, .05) will return .05 instead of 0 due to floor(). Using the aforementioned example, you may get 0 by replacing floor() with round() in a custom fmod().
<?
function fmod_round($x, $y) {
$i = round($x / $y);
return $x - $i * $y;
}
var_dump(fmod(.25, .05)); // float(0.05)
var_dump(fmod_round(.25, .05)); // float(0)
?>
cory at lavacube dot net ¶
19 years ago
I don't believe that is correct.
Try this out using your patch:
<?php
echo duration( mktime(0, 0, 0, 1, 0, 2006)-time() );
?>
As of right now, this will read:
1 month, 22 days, 24 hours, 49 minutes, 15 seconds
Which is completely incorrect. Seeing as how it is the 9th of December.
The real real flaw here is how the 'year' and 'month' periods are calculated. As most months vary in length...
Thank you very much SnakeEater251 for pointing this out.
The quickest way to get slightly more accurate results, is to use averages based on one "true" year, which is 365.25 days.
Change the year and month to:
'year' => 31557600, // one 'true year' (365.25 days)
'month' => 2629800, // one 'true year' divided by 12 :-)
I will work on developing a true fix, for pin-point accuracy. ;-)
- Cory Christison
dePijd ¶
15 years ago
This class ran through several unit tests and fixes all failures found in bugs.php.net
<?php
abstract class MyNumber {
public static function isZero($number, $precision = 0.0000000001)
{
$precision = abs($precision);
return -$precision < (float)$number && (float)$number < $precision;
}
public static function isEqual($number1, $number2)
{
return self::isZero($number1 - $number2);
}
public static function fmod($number1, $number2)
{
$rest = fmod($number1, $number2);
if (self::isEqual($rest, $number2)) {
return 0.0;
}
if (mb_strpos($number1, ".") === false) {
$decimals1 = 0;
} else {
$decimals1 = mb_strlen($number1) - mb_strpos($number1, ".") - 1;
}
if (mb_strpos($number2, ".") === false) {
$decimals2 = 0;
} else {
$decimals2 = mb_strlen($number2) - mb_strpos($number2, ".") - 1;
}
return (float)round($rest, max($decimals1, $decimals2));
}
}
?>
dan danschafer net ¶
6 years ago
WARNING: Due to how floating point numbers work, fmod() and any simple alternatives are problematic when there is either a massive orders of magnitude different between the input $x and $y, or the input and output values. If you need to work with large numbers or arbitrary precision, it is best to work with something like BC Math or GMP.
When working around fmod()'s problems, remember that floor() always goes towards -INF, not 0. This causes a commonly proposed fmod() alternative to only work with positive numbers:
<?php
function fmod_positive_only($x, $y) {
return $x - floor($x/$y) * $y;
}
?>
Given these simplistic input values:
fmod_positive_only(-5, 3) = 1 (wrong)
-5 % 3 = -2 (correct)
Correctly removing the decimal part of the quotient can be achieved with either casting to an int (always goes towards zero) or dynamically choosing ceil() or floor(). Dynamically choosing floor or ceil in an attempt to keep precision is overkill. If your $x and $y values are so different that it suffers from an overflow problem when casting, it was probably going to have precision problems anyway (see warnings below).
<?php
function fmod_overkill($x, $y) {
if (!$y) { return NAN; }
$q = $x / $y;
$f = ($q < 0 ? 'ceil' : 'floor');
return $x - $f($q) * $y;
}
?>
This is the "best" alternative for fmod() when given "normal" numbers.
<?php
function fmod_alt($x, $y) {
if (!$y) { return NAN; }
return floatval($x - intval($x / $y) * $y);
}
?>
WARNING: Even when you get a non-zero response, know your input numbers and when fmod() can go wrong. For large values or depending on your input variable types, float still may not contain enough precision to get back the correct answer. Here are a few problems with fmod() and their alternatives.
PHP_INT_MAX = 9223372036854775807
fmod(PHP_INT_MAX, 2) = 0 (wrong)
fmod_alt(PHP_INT_MAX, 2) = 0 (wrong)
PHP_INT_MAX % 2 = 1 (correct)
fmod(PHP_INT_MAX, PHP_INT_MAX - 1) = 0 (wrong)
fmod_alt(PHP_INT_MAX, PHP_INT_MAX - 1) = 1 (correct)
fmod_alt(PHP_INT_MAX, PHP_INT_MAX - 1.0) = 0 (wrong)
PHP_INT_MAX % (PHP_INT_MAX - 1) = 1 (correct)
PHP_INT_MAX % (PHP_INT_MAX - 1.0) = 9223372036854775807 (wrong)
fmod(PHP_INT_MAX, 131) = 98 (wrong)
fmod_alt(PHP_INT_MAX, 131) = 359 (wrong)
fmod_positive_only(PHP_INT_MAX, 131) = 0 (wrong)
PHP_INT_MAX % 131 = 97 (correct)
alex at xelam dot net ¶
21 years ago
Integer Modulo
If you want the remainder of the division of two Integers rather than Floats, use "%"; eg:
<?php
$a = 4;
$b = 3;
print($a % $b);
?>
Will output "1".
picaune at hotmail dot com ¶
21 years ago
NAN (.net Equivalent = Double.NaN) means "Not-a-Number".
Some ways to get NaN are modulo 0, and square root of 0.