I am new to php and was doing some mathematics and found this weird thing happening.
$numOfDiffChars = 62;
$id = 285355773910;
$remainder = $id % $numOfDiffChars;
echo "Remainder: ".$remainder." ID: ".$id." NumOfElems: ".$numOfDiffChars." ",($id - floor($id/$numOfDiffChars)*$numOfDiffChars);
The answer is as follows:
Remainder: 10 ID: 285355773910 NumOfElems: 62 26
which states that % operator gives the remainder 10 whereas mathematically its 26. What could be the reason for this? Is it just some error that I committed or is there a logic?
I am not quite sure, but if your using a 32-bit machine the problem could be that the integer $id is out of bound and therefore interpreted as a floating point.
http://php.net/manual/en/language.types.integer.php
have you tried fmod function find example on --->
http://www.php.net/manual/en/function.fmod.php
From php.net
Note that operator % (modulus) works just with integers (between -214748348 and 2147483647) while fmod() works with short and large numbers.
Related
Modulus operator is supposed to show the remainder. Like for echo(34%100) outputs 34. But why do i get a "Division by zero" error for this code echo(34%4294967296)
4294967296 is 2^32 and cannot be represented as 32 bit number - it wraps back to 0. If you use 64-bit version of PHP, it may work.
You might be able to use floating point modulus fmod to get what you want without overflowing.
https://bugs.php.net/bug.php?id=51731
2^31 is the largest integer you can get on Windows.
If you still want to mod large numbers, use bcmod.
I found this question searching for "division by zero error when using modulus", but the reason why was different.
Modulus (the % operator) will not work when the denomenator is is less than 1. using fmod() solves the problem.
Example:
$num = 5.1;
$den = .25;
echo ($num % $den);
// Outputs Warning: Division by zero
echo fmod($num, $den);
// Outputs 0.1
$num = 5.1;
$den = 1;
echo ($num % $den);
// Outputs 0, which is incorrect
echo fmod($num, $den);
// Outputs 0.1, which is correct
There's a lot of reports of mod getting wonky with large integers in php. Might be an overflow in the calculation or even in that number itself which is going to give you bugs. Best to use a large number library for that. Check out gmp or bcmath.
Does modulus division only return integers? I need a float return. See the following:
var_dump(12 % 10); // returns 2, as expected
var_dump(11.5 % 10); // returns 1 instead of 1.5?
Yes. the % operator returns an integer.
If you want a floating point result, use the fmod() function instead.
See the manual.
Operands of modulus are converted to integers (by stripping the
decimal part) before processing.
11.5 becomes 11.
11 % 10 = 1 remainder **1**
Your solution: fmod(), as tom_yes_tom suggests.
Quoting the documentation page:
"Operands of modulus are converted to integers (by stripping the
decimal part) before processing."
Is there any workaround for this?
mathematics...
11.5 - floor(11.5 / 10) * 10 == 1.5
I've been wrestling with PHP's ceil() function giving me slightly wrong results - consider the following:
$num = 2.7*3; //float(8.1)
$num*=10; //float(81)
$num = ceil($num); //82, but shouldn't this be 81??
$num/=10; //float(8.2)
I have a number which may have any number of decimal places, and I need it rounded up to one decimal place.
i.e 8.1 should be 8.1, 8.154 should be 8.2, and 8 should be left as 8.
How I've been getting there is to take the number, multiply by 10, ceil() it, then divide by ten but as you can see I'm getting an extra .1 added in some circumstances.
Can anyone tell my why this is happening, and how to fix it?
Any help greatly appreciated
EDIT: had +=10 instead of *=10 :S
EDIT 2:
I didn't explicitly mention this but I need the decimal to ALWAYS round UP, never down - this answer is closest so far:
rtrim(rtrim(sprintf('%.1f', $num), '0'), '.');
However rounds 3.84 down to 3.8 when I need 3.9.
Sorry this wasn't clearer :(
Final Edit:
What I ended up doing was this:
$num = 2.7*3; //float(8.1)
$num*=10; //float(81)
$num = ceil(round($num, 2)); //81 :)
$num/=10; //float(8.1)
Which works :)
This is more than likely due to floating point error.
http://support.microsoft.com/kb/42980
http://download.oracle.com/docs/cd/E19957-01/806-3568/ncg_goldberg.html
http://joshblog.net/2007/01/30/flash-floating-point-number-errors/
http://en.wikipedia.org/wiki/Floating_point
You may have luck trying this procedure instead.
<?php
$num = 2.7*3;
echo rtrim(rtrim(sprintf('%.1f', $num), '0'), '.');
Floats can be a fickle thing. Not all real numbers can be properly represented in a finite number of binary bits.
As it turns out, a decimal section of 0.7 is one of those numbers (comes out 0.10 with an infinity repeating "1100" after it). You end up with a number that's ever so slightly above 0.7, so when you multiply by 10, you have a one's digit slightly above 7.
What you can do is make a sanity check. Take you float digit and subtract it's integer form. If the resulting value is less than, say, 0.0001, consider it to be an internal rounding error and leave it as-is. If the result is greater than 0.0001, apply ceil() normally.
Edit: A fun example you can do if you're on windows to show this is to open up the built in calculator application. Put in "4" then apply a square root function (with x^y where y=0.5). You'll see it properly displays "2". Now, subtract 2 from it and you'll see that you don't have 0 as a result. This is caused by internal rounding errors when it attempted to compute the square root of 4. When displaying the number 2 earlier, it knew that those very distant trailing digits were probably a rounding error, but when those are all that's left, it gets a bit confused.
(Before anybody gets onto me about this, I understand that this is oversimplified, but nonetheless I consider it a decent example.)
Convert your number to a string and ceil the string.
function roundUp($number, $decimalPlaces){
$multi = pow(10, $decimalPlaces);
$nrAsStr = ($number * $multi) . "";
return ceil($nrAsStr) / $multi;
}
The problem is that floating point numbers are RARELY what you expect them to be. Your 2.7*3 is probably coming out to be something like 81.0000000000000000001, which ceil()'s up to 82. For this sort of thing, you'll have to wrap your ceil/round/floor calls with some precision checks, to handle those extra microscopic differences.
Use %f instead of %.1f.
echo rtrim(rtrim(sprintf('%f', $num), '0'), '.');
Why not try this:
$num = 2.7*3;
$num *= 100;
$num = floor($num);
$num /= 10;
$num = ceil($num);
$num /= 10;
I observed the following and would be thankful for an explanation.
$amount = 4.56;
echo ($amount * 100) % 5;
outputs : 0
However,
$amount = 456;
echo $amount % 5;
outputs : 1
I tried this code on two separate PHP installations, with the same result. Thanks for your help!
I strongly suspect this is because 4.56 can't be exactly represented as a binary floating point number, so a value very close to it is used instead. When multiplied by 100, that comes to 455.999(something), and then the modulo operator truncates down to 455 before performing the operation.
I don't know the exact details of PHP floating point numbers, but the closest IEEE-754 double to 4.56 is 4.55999999999999960920149533194489777088165283203125.
So here's something to try:
$amount = 455.999999999;
echo $amount % 5;
I strongly suspect that will print 0 too. From some PHP arithmetic documentation:
Operands of modulus are converted to integers (by stripping the decimal part) before processing.
Use fmod to avoid this problem.
Derived from this question : (Java) How does java do modulus calculations with negative numbers?
Anywhere to force PHP to return positive 51?
update
Looking for a configuration setting to fix, instead hard-guessing
Or other math function like bcmath?
updated
Not entire convinced by that java answer, as it does not take account of negative modulus
-13+(-64) =?
Anyway, the post you referenced already gave the correct answer:
$r = $x % $n;
if ($r < 0)
{
$r += abs($n);
}
Where $x = -13 and $n = 64.
If GMP is available, you can use gmp_mod
Calculates n modulo d. The result is always non-negative, the sign of d is ignored.
Example:
echo gmp_strval(gmp_mod('-13', '64')); // 51
Note that n and d have to be GMP number resources or numeric strings. Anything else won't work¹
echo gmp_strval(gmp_mod(-13, 64));
echo gmp_mod(-13, 64);
will both return -51 instead (which is a bug).
¹ running the above in this codepad, will produce 51 in all three cases. It won't do that on my development machine.
The modulo operation should find the remainder of division of a number by another. But strictly speaking in most mainstream programming languages the modulo operation malfunctions if dividend or/and divisor are negative. This includes PHP, Perl, Python, Java, C, C++, etc.
Why I say malfunction? Because according to mathematic definition, a remainder must be zero or positive.
The simple solution is to handle the case yourself:
if r < 0 then r = r + |divisor|;
|divisor| is the absolute value of divisor.
Another solution is to use a library (as #Gordon pointed). However I wouldn't use a library to handle a simple case like this.
I hate using if in this case when you can calculate it right away.
$r = ($x % $n + $n) % $n;
when $n is positive.
The PHP manual says that
The result of the modulus operator % has the same sign as the dividend — that is, the result of $a % $b will have the same sign as $a. For example
so this is not configurable. Use the options suggested in the question you linked to