This problem is best expressed in code:
$var1 = 286.46; // user input data
$var2 = 3646; // user input data
$var3 = 25000; // minumum amount allowed
$var4 = ($var1 * 100) - $var2; // = 250000
if ($var4 < $var3) { // if 250000 < 250000
print 'This returns!';
}
var_dump($var4) outputs: float(25000) and when cast to int, outputs: int(24999) - and thereby lies the problem.
I don't really know what to do about it though. The issue occurs upon multiplication by 100, and while there are little tricks I can do to get around that (such as *10*10) I'd like to know if there's a 'real' solution to this problem.
Thanks :)
This is a horrible hacky solution and I slightly hate myself for it, but this gives the expected behaviour:
<?php
$var1 = 286.46; // user input data
$var2 = 3646; // user input data
$var3 = 25000; // minumum amount allowed
$var4 = ($var1 * 100) - $var2; // = 250000
if ((string) $var4 < (string) $var3) { // if 250000 < 250000
print 'This returns!';
}
Cast them to strings, and they get converted back to int/float as appropriate for the comparison. I don't like it but it does work.
Really you need BC Math for precise floating point mathematics in PHP.
Its always a good idea to use ceil (or floor based on what you want) when using float number as int
In your case try ceil($var4) before comparison!
That's what floats do sometimes, it is all due to how floats are unable to precisely represent integers from time to time.
Instead of casting it to an int, you can round the number to an integer value and then cast it to an int. (possibly that cast unnecessary, but PHP isn't to clear about how such things happen internally, and even if you know how they happen right now, they may not in the future.
I think you could use bccomp for comparing floating point values but i think it's a function that's not in the PHP Core.
Otherwise i found this function here but i couldn't test it to see if it works
function Comp($Num1,$Num2,$Scale=null) {
// check if they're valid positive numbers, extract the whole numbers and decimals
if(!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num1,$Tmp1)||
!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num2,$Tmp2)) return('0');
// remove leading zeroes from whole numbers
$Num1=ltrim($Tmp1[1],'0');
$Num2=ltrim($Tmp2[1],'0');
// first, we can just check the lengths of the numbers, this can help save processing time
// if $Num1 is longer than $Num2, return 1.. vice versa with the next step.
if(strlen($Num1)>strlen($Num2)) return(1);
else {
if(strlen($Num1)<strlen($Num2)) return(-1);
// if the two numbers are of equal length, we check digit-by-digit
else {
// remove ending zeroes from decimals and remove point
$Dec1=isset($Tmp1[2])?rtrim(substr($Tmp1[2],1),'0'):'';
$Dec2=isset($Tmp2[2])?rtrim(substr($Tmp2[2],1),'0'):'';
// if the user defined $Scale, then make sure we use that only
if($Scale!=null) {
$Dec1=substr($Dec1,0,$Scale);
$Dec2=substr($Dec2,0,$Scale);
}
// calculate the longest length of decimals
$DLen=max(strlen($Dec1),strlen($Dec2));
// append the padded decimals onto the end of the whole numbers
$Num1.=str_pad($Dec1,$DLen,'0');
$Num2.=str_pad($Dec2,$DLen,'0');
// check digit-by-digit, if they have a difference, return 1 or -1 (greater/lower than)
for($i=0;$i<strlen($Num1);$i++) {
if((int)$Num1{$i}>(int)$Num2{$i}) return(1);
else
if((int)$Num1{$i}<(int)$Num2{$i}) return(-1);
}
// if the two numbers have no difference (they're the same).. return 0
return(0);
}
}
}
The problem is that floats just cannot represent some numbers. Since PHP doesn't have a "decimal" (or other fixed-point) type, you can basically only hack your way around these problems.
Assuming the first number in your example $var1 = 286.46 denotes some kind of money, you could just convert that to cents directly after the user entered it (e.g. through stripping the point and reading it as an integer) and thus calculate everything using integer math.
That's not a general solution - and I doubt that one exists (short of using arbitrary precision numbers, which some PHP extensions provide - but I that smells like overkill to me).
Related
I have two variables in a PHP program for billing statements, $charges and $payments.
$charges is the total amount due before any payments. $payments is the total amount received.
I calculate the balance due like so:
$balance_due = $charges-$payments;
Simple, except I am getting the following result:
$balance_due has -9.0949470177293E-13 for a value (expecting 0).
Both $charges and $payments have a value of 5511.53.
When I var_dump($charges) and var_dump($payments) they both show: float(5511.53)
This code (and === ):
if($charges == $payments){
error_log('they are the same');
}else{
error_log('they are not the same');
}
both result in false.
If I hard code: $charges = $payments = 5511.53; and run it then $balance_due = 0 as expected.
I am confused. What am I missing?
EDIT NOTES
I was able to use a user contributed function by Nitrogen found on the BC Math Functions page that was suggested I look at in order to come up with the following solution:
if(Comp($charges, $payments)===0){
$balance_due = 0;
}else{
$balance_due = ( $charges - $payments );
}
function Comp($Num1,$Num2,$Scale=null) {
// check if they're valid positive numbers, extract the whole numbers and decimals
if(!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num1,$Tmp1)||
!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num2,$Tmp2)) return('0');
// remove leading zeroes from whole numbers
$Num1=ltrim($Tmp1[1],'0');
$Num2=ltrim($Tmp2[1],'0');
// first, we can just check the lengths of the numbers, this can help save processing time
// if $Num1 is longer than $Num2, return 1.. vice versa with the next step.
if(strlen($Num1)>strlen($Num2)) return(1);
else {
if(strlen($Num1)<strlen($Num2)) return(-1);
// if the two numbers are of equal length, we check digit-by-digit
else {
// remove ending zeroes from decimals and remove point
$Dec1=isset($Tmp1[2])?rtrim(substr($Tmp1[2],1),'0'):'';
$Dec2=isset($Tmp2[2])?rtrim(substr($Tmp2[2],1),'0'):'';
// if the user defined $Scale, then make sure we use that only
if($Scale!=null) {
$Dec1=substr($Dec1,0,$Scale);
$Dec2=substr($Dec2,0,$Scale);
}
// calculate the longest length of decimals
$DLen=max(strlen($Dec1),strlen($Dec2));
// append the padded decimals onto the end of the whole numbers
$Num1.=str_pad($Dec1,$DLen,'0');
$Num2.=str_pad($Dec2,$DLen,'0');
// check digit-by-digit, if they have a difference, return 1 or -1 (greater/lower than)
for($i=0;$i<strlen($Num1);$i++) {
if((int)$Num1{$i}>(int)$Num2{$i}) return(1);
else
if((int)$Num1{$i}<(int)$Num2{$i}) return(-1);
}
// if the two numbers have no difference (they're the same).. return 0
return(0);
}
}
}
That solution worked for me. The answer provided by imtheman below also works and seems more efficient so I am going to use that one instead. Is there any reason not to use one or the other of these?
The way I solved this problem when I ran into it was using php's number_format(). From php documentation:
string number_format(float $number [, int $decimals = 0 ])
So what I would do is this:
$balance_due = number_format($charges-$payments, 2);
And that should solve your problem.
Note: number_format() will return a string, so to compare it you must use == (not ===) or cast it back into a (float) before comparison.
I've got this spot of code that seems it could be done cleaner with pure math (perhaps a logarigthms?). Can you help me out?
The code finds the first power of 2 greater than a given input. For example, if you give it 500, it returns 9, because 2^9 = 512 > 500. 2^8 = 256, would be too small because it's less than 500.
function getFactor($iMaxElementsPerDir)
{
$aFactors = range(128, 1);
foreach($aFactors as $i => $iFactor)
if($iMaxElementsPerDir > pow(2, $iFactor) - 1)
break;
if($i == 0)
return false;
return $aFactors[$i - 1];
}
The following holds true
getFactor(500) = 9
getFactor(1000) = 10
getFactor(2500) = 12
getFactor(5000) = 13
You can get the same effect by shifting the bits in the input to the right and checking against 0. Something like this.
i = 1
while((input >> i) != 0)
i++
return i
The same as jack but shorter. Log with base 2 is the reverse function of 2^x.
echo ceil(log(500, 2));
If you're looking for a "math only" solution (that is a single expression or formula), you can use log() and then take the ceiling value of its result:
$factors = ceil(log(500) / log(2)); // 9
$factors = ceil(log(5000) / log(2)); // 13
I seem to have not noticed that this function accepts a second argument (since PHP 4.3) with which you can specify the base; though internally the same operation is performed, it does indeed make the code shorter:
$factors = ceil(log(500, 2)); // 9
To factor in some inaccuracies, you may need some tweaking:
$factors = floor(log($nr - 1, 2)) + 1;
There are a few ways to do this.
Zero all but the most significant bit of the number, maybe like this:
while (x & x-1) x &= x-1;
and look the answer up in a table. Use a table of length 67 and mod your power of two by 67.
Binary search for the high bit.
If you're working with a floating-point number, inspect the exponent field. This field contains 1023 plus your answer, except in the case where the number is a perfect power of two. You can detect the perfect power case by checking whether the significand field is exactly zero.
If you aren't working with a floating-point number, convert it to floating-point and look at the exponent like in 3. Check for a power of two by testing (x & x-1) == 0 instead of looking at the significand; this is true exactly when x is a power of two.
Note that log(2^100) is the same double as log(nextafter(2^100, 1.0/0.0)), so any solution based on floating-point natural logarithms will fail.
Here's (nonconformant C++, not PHP) code for 4:
int ceillog2(unsigned long long x) {
if (x < 2) return x-1;
double d = x-1;
int ans = (long long &)d >> 52;
return ans - 1022;
}
I have the following function that determines if I sale is fully paid for. I don't remember why I did it this way, but it has been working so far and I don't remember why I had to do it this way.
function _payments_cover_total()
{
//get_payments is a list of payment amounts such as:
//10.20, 10.21, or even 10.1010101101 (10 decimals max)
$total_payments = 0;
foreach($this->sale_lib->get_payments() as $payment)
{
$total_payments += $payment['payment_amount'];
}
//to_currency_no_money rounds total to 2 decimal places
if (to_currency_no_money($this->sale_lib->get_total()) - $total_payments ) > 1e-6 ) )
{
return false;
}
return true;
}
I am wondering if there is ever a case where due to a rounding error that this function would return false when it shouldn't.
The main part I have a question about is:
> 1e-6
I think before I had, but it was causing problems in some cases.
> 0
I think you are doing what is mentioned on php floating help page. To quote it directly :
To test floating point values for equality, an upper bound on the
relative error due to rounding is used. This value is known as the
machine epsilon, or unit roundoff, and is the smallest acceptable
difference in calculations.
$a and $b are equal to 5 digits of precision.
<?php
$a = 1.23456789;
$b = 1.23456780;
$epsilon = 0.00001;
if(abs($a-$b) < $epsilon) {
echo "true";
}
?>
So in your case:
(to_currency_no_money($this->sale_lib->get_total()) - $total_payments) > 1e-6
relative error due to rounding should not be great than 1e-6 or 0.000001
if you are not sure about left operand being greater than right 100% time,then you should add abs() e.g for correctness.
$relative_error=to_currency_no_money($this->sale_lib->get_total()) - $total_payments;
if(abs($relative_error) > 1e-6){
return false
}
return true;
$x = (1.333-1.233)-(1.334-1.234);
echo $x;
//result = $x = -2.2204460492503E-16 - close to zero
//but (1.333-1.233)-(1.334-1.234) = 0.1 - 0.1 = 0 (in calculator)
if($x === 0){
echo "|zero";
}
else {
echo "|non zero"; //<== this is result
}
//screen = -2.2204460492503E-16|non zero
//how to get to zero?
if($x > 1e-6){//1e-6 mathematical constant
echo "|non zero";
}
else {
echo "|zero";//this is result
}
//screen -2.2204460492503E-16|non zero|zero
if ($x > 1e-6 )
{
echo " false";
//echo "|non zero";
//return false;
}
else{
echo " true";//<== this resut
//echo "|zero";
//return true;
}
//screen -2.2204460492503E-16|non zero|zero true
printf("%.1f<br />", 1e-1);
printf("%.2f<br />", 1e-2);
printf("%.3f<br />", 1e-3);
printf("%.4f<br />", 1e-4);
printf("%.5f<br />", 1e-5);
printf("%.6f<br />", 1e-6);
printf("%.7f<br />", 1e-7);
printf("%.8f<br />", 1e-8);
printf("%.9f<br />", 1e-9);
printf("%.10f<br />", 1e-10);
printf("%.11f<br />", 1e-11);
printf("%.12f<br />", 1e-12);
printf("%.29f<br />", -2.2204460492503E-16);
//0.1
//0.01
//0.001
//0.0001
//0.00001
//0.000001
//0.0000001
//0.00000001
//0.000000001
//0.0000000001
//0.00000000001
//0.000000000001
//-0.00000000000000022204460492503
I am sorry, but when dealing with currency, you shouldn't really be using PHP floats, as IMSoP stated. The reason is also from PHP float help pages:
Additionally, rational numbers that are exactly representable as
floating point numbers in base 10, like 0.1 or 0.7, do not have an
exact representation as floating point numbers in base 2, which is
used internally, no matter the size of the mantissa. Hence, they
cannot be converted into their internal binary counterparts without a
small loss of precision. This can lead to confusing results: for
example, floor((0.1+0.7)*10) will usually return 7 instead of the
expected 8, since the internal representation will be something like
7.9999999999999991118....
So never trust floating number results to the last digit, and do not
compare floating point numbers directly for equality. If higher
precision is necessary, the arbitrary precision math functions and gmp
functions are available.
Note that the help page specifically says you can't trust float results to the last digit, no matter how short (after comma) it is.
So while you do have very short floats (just 2 digits after comma), the float precision (1e-6) doesn't enter into it really, and you can't trust them 100%.
Since it is a question of money, in order to avoid angry customers and lawsuits accusing of penny shaving (https://en.wikipedia.org/wiki/Salami_slicing), the real solutions are:
1) either use PHP BC math, which works with string representation of numbers
http://php.net/manual/en/book.bc.php
2) as IMSoP suggested, use integers and store the amounts in smallest denomination (cents, eurocents or whatever you have) internally.
First solution might be a bit more resource intense: I haven't used BC math myself much, but storing strings and doing arbitrary precision math (which might be a bit of an overkill in this case) are by definition more RAM and CPU intense than working with integers.
It might, however, need less changes in other parts of the code.
The second solution requires changes to represenation, so that wherever user sees the amounts, they are normalized to dollars,cents (or whatever have you).
Note, however, that in this case also you run problems with rounding risks at some point, as when you do:
float shown_amount; // what we show to customer: in dollars,cents
int real_amount; // what we use internally: in cents
shown_amount = cent_amount / 100;
you may reintroduce the rounding problems as you have floats again and possibilities for rounding errors, so tread carefully and be sure to make calculations and roundings only on real_amount, never on shown_amount
this might be a stupid question but I have searched again and again without finding any results.
So, what I want is to show all the decimal places of a number without knowing how many decimal places it will have. Take a look at this small code:
$arrayTest = array(0.123456789, 0.0123456789);
foreach($arrayTest as $output){
$newNumber = $output/1000;
echo $newNumber;
echo "<br>";
}
It gives this output:
0.000123456789
1.23456789E-5
Now, I tried using 'number_format', but I don't think that is a good solution. It determines an exact amount of decimal places, and I do not know the amount of decimal places for every number. Take a look at the below code:
$arrayTest = array(0.123456789, 0.0123456789);
foreach($arrayTest as $output){
$newNumber = $output/1000;
echo number_format($newNumber,13);
echo "<br>";
}
It gives this output:
0.0001234567890
0.0000123456789
Now, as you can see there is an excess 0 in the first number, because number_format forces it to have 13 decimal places.
I would really love some guidance on how to get around this problem. Is there a setting in PHP.ini which determines the amount of decimals?
Thank you very much in advance!
(and feel free to ask if you have any further questions)
It is "impossible" to answer this question properly - because a binary float representation of a decimal number is approximate: "What every computer scientist should know about floating point"
The closest you can come is write yourself a routine that looks at a decimal representation of a number, and compares it to the "exact" value; once the difference becomes "small enough for your purpose", you stop adding more digits.
This routine could then return the "correct number of digits" as a string.
Example:
<?php
$a = 1.234567890;
$b = 0.123456789;
echo returnString($a)."\n";
echo returnString($b)."\n";
function returnString($a) {
// return the value $a as a string
// with enough digits to be "accurate" - that is, the value returned
// matches the value given to 1E-10
// there is a limit of 10 digits to cope with unexpected inputs
// and prevent an infinite loop
$conv_a = 0;
$digits=0;
while(abs($a - $conv_a) > 1e-10) {
$digits = $digits + 1;
$conv_a = 0 + number_format($a, $digits);
if($digits > 10) $conv_a = $a;
}
return $conv_a;
}
?>
Which produces
1.23456789
0.123456789
In the above code I arbitrarily assumed that being right to within 1E-10 was good enough. Obviously you can change this condition to whatever is appropriate for the numbers you encounter - and you could even make it an optional argument of your function.
Play with it - ask questions if this is not clear.
See, I want to write a function that takes a float number parameter and rounds the float to the nearest currency value (a float with two decimal places) but if the float parameter has a zero fraction (that is, all zeroes behind the decimal place) then it returns the float as an integer (or i.e. truncates the decimal part since they're all zeroes anyways.).
However, I'm finding that I can't figure out how to determine if if a fraction has a zero fraction. I don't know if there's a PHP function that already does this. I've looked. The best I can think of is to convert the float number into an integer by casting it first and then subtract the integer part from the float and then check if the difference equals to zero or not.
if($value == round($value))
{
//no decimal, go ahead and truncate.
}
This example compares the value to itself, rounded to 0 decimal places. If the value rounded is the same as the value, you've got no decimal fraction. Plain and simple.
A little trick with PHPs type juggling abilities
if ($a == (int) $a) {
// $a has a zero fraction value
}
I think the best way:
if ((string)$value == (int)$value){
...
}
Example:
$value = 2.22 * 100;
var_dump($value == (int)$value); // false - WRONG!
var_dump($value == round($value)); // false - WRONG!
var_dump((string)$value == (int)$value); // true - OK!
function whatyouneed($number) {
$decimals = 2;
printf("%.".($number == (int)($number) ? '0' : $decimals)."F", $number);
}
So basically it's either printf("%.2F") if you want 2 decimals and printf("%.2F") if you want none.
Well, the problem is that floats aren't exact. Read here if you're interested in finding out why. What I would do is decide on a level of accuracy, for example, 3 decimal places, and base exactness on that. To do that, you multiply it by 1000, cast it to an int, and then check if $your_number % 1000==0.
$mynumber = round($mynumber *1000);
if ($mynumber % 1000==0)
{ isInt() }
Just so you know, you don't have to write a function to do that, there's already one that exists:
$roundedFloat = (float)number_format("1234.1264", 2, ".", ""); // 1234.13
If you want to keep the trailing .00, just omit the float cast (although it will return a string):
$roundedFloatStr = number_format("1234.000", 2, ".", ""); // 1234.00