phpday 2022

fmod

(PHP 4 >= 4.2.0, PHP 5, PHP 7, PHP 8)

fmodВозвращает дробный остаток от деления по модулю

Описание

fmod(float \$num1, float \$num2): float

Возвращает дробный остаток от деления делимого (num1) и делителя (num2). Остаток (r) определяется так: x = i * y + r, где i - некоторое целое. Если y не равен нулю, то r имеет такой же знак, как num1 и величину, меньшую или равную величине num2.

num1

Делимое.

num2

Делитель.

Возвращаемые значения

Остаток от операции деления num1/num2

Примеры

Пример #1 Пример использования fmod()

<?php
\$x
5.7;
\$y 1.3;
\$r fmod(\$x\$y);
// \$r равно 0.5, потому что 4 * 1.3 + 0.5 = 5.7
?>

Смотрите также

• / - деление с плавающей точкой
• % - целочисленный остаток
• intdiv() - целочисленное деление

jphansen at uga dot edu
17 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)
?>
dan danschafer net
3 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)
cory at lavacube dot net
16 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
ysangkok at gmail dot com
14 years ago
<?php
function custom_modulo(\$var1, \$var2) {

\$tmp = \$var1/\$var2;

return (float) (
\$var1 - ( ( (int) (\$tmp) ) * \$var2 ) );
}

\$par1 = 1;
\$par2 = 0.2;

echo
"fmod:          ";
var_dump(fmod ( \$par1 , \$par2 ));
echo
"custom_modulo: ";
var_dump(custom_modulo ( \$par1 , \$par2 ));
?>

gives this:

fmod:          float(0.2)
custom_modulo: float(0)

Fmod does not deliver the desired result, therefore I made my own.
picaune at hotmail dot com
18 years ago
NAN (.net Equivalent = Double.NaN) means "Not-a-Number".
Some ways to get NaN are modulo 0, and square root of 0.
dePijd
11 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));
}
}
?>
alex at xelam dot net
17 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".
cory at simplesystems dot ca
16 years ago
Just a note on the previous note by Ryan Means:

Instead of using explode() to get the number before the decimal point, would be to use floor()... floor() rounds fractions down, which is exactly what is needed.

His same example using floor();

<?PHP
\$totalsec
=XXXXXXX; //Replace the X's with a int value of seconds

\$daysarray = floor( \$totalsec/86400 );

\$partdays = fmod(\$totalsec, 86400);
\$hours = floor( \$partdays/3600 );

\$parthours = fmod(\$partdays, 3600);
\$min = floor( \$parthours/60 );

\$sec = fmod(\$parthours, 60);

echo
"days " . \$days . "<br>";
echo
"hours " . \$hours . "<br>";
echo
"minutes " . \$min . "<br>";
echo
"seconds " . \$sec . "<br>";
?>
rocan
16 years ago
john at digitizelife dot com:

Well not sure how your comment applys to fmod..

but their is a sure simpler way of coping with situations like this..

its called a bit field (bit masking)

e.g.

/* Categories */
bin     dec   cat
0001 - 1 - Blue
0010 - 2 - Red
0100 - 4 - Green
1000 - 8 - Yellow

/* Permissions */
0010 - 2   - Bob
0101 - 5    - John
1011 - 11  - Steve
1111-  15 - Mary

to find out the permissions for each user you simple need to do a bitwise AND

\$steve_auth=11;

function get_perm(\$auth)
{
\$cats["Blue"]=1;
\$cats["Red"]=2;
\$cats["Green"]=4;
\$cats["Yellow"]=8;
\$perms=array();
foreach(\$cats as \$perm=>\$catNum)
{
if(\$auth & \$catNum)
\$perms[\$perm]=true;

}

return \$perms;
}

print_r(get_perm(\$steve_auth));
/*
returns
Array
(
[Blue] => 1
[Red] => 1
[Yellow] => 1
)
*/

This is far simpler than your prime number idea, in fact you dont even need a function in any tests for the permmsions on a user you can do them directly using the bitwise and operator.

You may want to read the following

http://uk2.php.net/manual/en/language.operators.bitwise.php
konstantin at rekk dot de
17 years ago
If you need to reduce an integer to zero if zero and 1 if not, you can use

\$sign = (integer)(boolean)\$integer;

\$sign = \$integer > 0 ? 1 : 0;

it is faster from 100 operations on (at least on my machine).
cory at lavacube dot net
16 years ago
A more formal way for generating duration strings:

<?php

function duration( \$int_seconds=0, \$if_reached=null )
{

\$key_suffix = 's';

\$periods = array(

'year'        => 31556926,

'month'        => 2629743,

'day'        => 86400,

'hour'        => 3600,

'minute'    => 60,

'second'    => 1

);

// used to hide 0's in higher periods

\$flag_hide_zero = true;

// do the loop thang

foreach( \$periods as \$key => \$length )
{

// calculate

\$temp = floor( \$int_seconds / \$length );

// determine if temp qualifies to be passed to output

if( !\$flag_hide_zero || \$temp > 0 )
{

// store in an array

\$build[] = \$temp.' '.\$key.(\$temp!=1?'s':null);

// set flag to false, to allow 0's in lower periods

\$flag_hide_zero = false;
}

// get the remainder of seconds

\$int_seconds = fmod(\$int_seconds, \$length);
}

// return output, if !empty, implode into string, else output \$if_reached

return ( !empty(\$build)?implode(', ', \$build):\$if_reached );
}

?>

Simple use:
<?php

echo duration( mktime(0, 0, 0, 1, 1, date('Y')+1) - time(), 'Some fancy message to output if duration is already met...' );

?>

Enjoy. :-)
KRAER
7 years ago
To create a list of primes in a bash based on php wich can be resumed after breaking I did use fmod() and some snippets offered by two more users here on php comments.

This will output :
"prime;difference-between-last-and-current-prime"

So credit goes to them. I only did the logfile output.

This will function up to whatever fmod supports as highest value. Just enter the \$end value. And do a touch to the logfile followed by chmod 666 so php can access it.

<?php
function tailCustom(\$filepath, \$lines = 1, \$adaptive = true) {

// Open file

\$f = @fopen(\$filepath, "rb");
if (
\$f === false) return false;

// Sets buffer size

else
\$buffer = (\$lines < 2 ? 64 : (\$lines < 10 ? 512 : 4096));

fseek(\$f, -1, SEEK_END);

// (Otherwise the result would be wrong if file doesn't end with a blank line)

if (fread(\$f, 1) != "\n") \$lines -= 1;

\$output = '';

\$chunk = '';

// While we would like more

while (ftell(\$f) > 0 && \$lines >= 0) {

// Figure out how far back we should jump

\$seek = min(ftell(\$f), \$buffer);

// Do the jump (backwards, relative to where we are)

fseek(\$f, -\$seek, SEEK_CUR);

// Read a chunk and prepend it to our output

\$output = (\$chunk = fread(\$f, \$seek)) . \$output;

// Jump back to where we started reading

fseek(\$f, -mb_strlen(\$chunk, '8bit'), SEEK_CUR);

// Decrease our line counter

\$lines -= substr_count(\$chunk, "\n");

}

// While we have too many lines
// (Because of buffer size we might have read too many)

while (\$lines++ < 0) {

// Find first newline and remove all text before that

\$output = substr(\$output, strpos(\$output, "\n") + 1);

}

// Close file and return

fclose(\$f);
return
trim(\$output);

}

function
isPrime( \$num )
{
for(
\$i = 2; \$i*\$i <= \$num; \$i++ )
if( !
fmod(\$num,\$i) )
return
FALSE;

return
TRUE;
}

\$logfile = 'prim_save.log';

\$lastline = explode(";", tailCustom(\$logfile));
\$begin = (\$lastline +1);
\$lastprime = \$lastline;

\$end = 999999999999999999999999999999999999;

\$fp = fopen(\$logfile, 'a');
//Lineformat    \$i.';'.\$difference.';'."\n"

for(\$i = \$begin; \$i<\$end; \$i++)
{
if(
isPrime(\$i) == TRUE)
{

\$difference = \$i - \$lastprime;

fputs(\$fp,\$i.';'.\$difference.';'."\n");

\$lastprime = \$i;
}
}

fclose(\$fp);
?>
8 years ago
There is an elegant way to do compute gcm :
https://en.wikipedia.org/wiki/Greatest_common_divisor

// Recursive function to compute gcd (euclidian method)
function gcd (\$a, \$b) {
return \$b ? gcd(\$b, \$a % \$b) : \$a;
}
// Then reduce any list of integer
echo array_reduce(array(42, 56, 28), 'gcd'); // === 14

If you want to work with floating points, use approximation :

function fgcd (\$a, \$b) {
return \$b > .01 ? fgcd(\$b, fmod(\$a, \$b)) : \$a; // using fmod
}
echo array_reduce(array(2.468, 3.7, 6.1699), 'fgcd'); // ~= 1.232

You can use a closure in PHP 5.3 :

\$gcd = function (\$a, \$b) use (&\$gcd) { return \$b ? \$gcd(\$b, \$a % \$b) : \$a; };
matrebatre
13 years ago
I always use this:

function modulo(\$n,\$b) {
return \$n-\$b*floor(\$n/\$b);
}

And it appears to work correctly.
SnakeEater251
16 years ago
Note on the code given by cory at lavacube dot net.
You will recieve better results by not using floor and using round instead. As you continue increasing to larger amounts of time you will notice that the outputted time is off by large amounts.

so instead of \$temp = floor( \$int_seconds / \$length );
we would use  \$temp = round( \$int_seconds / \$length );

<?php

function duration( \$int_seconds=0, \$if_reached=null )
{

\$key_suffix = 's';

\$periods = array(

'year'        => 31556926,

'month'        => 2629743,

'day'        => 86400,

'hour'        => 3600,

'minute'    => 60,

'second'    => 1

);

// used to hide 0's in higher periods

\$flag_hide_zero = true;

// do the loop thang

foreach( \$periods as \$key => \$length )
{

// calculate

\$temp = round( \$int_seconds / \$length );

// determine if temp qualifies to be passed to output

if( !\$flag_hide_zero || \$temp > 0 )
{

// store in an array

\$build[] = \$temp.' '.\$key.(\$temp!=1?'s':null);

// set flag to false, to allow 0's in lower periods

\$flag_hide_zero = false;
}

// get the remainder of seconds

\$int_seconds = fmod(\$int_seconds, \$length);
}

// return output, if !empty, implode into string, else output \$if_reached

return ( !empty(\$build)?implode(', ', \$build):\$if_reached );
}

?>
radoslaw dot roszkowski at gmail dot com
3 years ago
Do not relly on that function, for example:

\$a = "7.191"
\$b =  "2.397000"

if(fmod(floatval(\$a), floatval(\$b)) === 0.0) {
..
//is false, becouse
//float(4.4408920985006E-16) != 0.0
nospam at neonit dot de
5 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.
timo underscore teichert at yahoo dot de
7 years ago
The behaviour of this function seems to have changed over time.

<?php

echo fmod(3,5);
// php 5.3.2 outputs -2
// php 5.3.8 outputs 3

echo fmod(2,5);
// php 5.3.2 outputs 2
// php 5.3.8 outputs 2

?>

- Timo
verdy_p
9 years ago
Note that fmod is NOT equivalent to this basic function:

<?php
function modulo(\$a, \$b) {
return
\$a - \$b * floor(\$a / \$b);
}
?>

because fmod() will return a value with the same sign as \$a. In other words the floor() function is not correct as it rounds towards -INF instead of towards zero.

To emulate fmod(\$a, \$b) the correct way is:

<?php
function fmod(\$a, \$b) {
return
\$a - \$b * ((\$b < 0) ? ceil(\$a / \$b) : floor(\$a / \$b)));
}
?>

Note that both functions will throw a DIVISION BY ZERO if \$b is null.

The first function modulo() above is the mathematical function which is useful for working on cyclic structures (such as calender computions or trignonometric functions :

- fmod(\$a, 2*PI) returns a value in [0..2*PI) if \$a is positive
- fmod(\$a, 2*PI) returns a value in [-2*PI..0] if \$a is negative
- modulo(\$a, 2*PI) returns a value always in [0..2*PI) independantly of the sign of \$a
-1 