PHP - Floating Number Precision

Question

$a = '35';
$b = '-34.99';
echo ($a + $b);

Results in 0.009999999999998

What is up with that? I wondered why my program kept reporting odd results.

Why doesn't PHP return the expected 0.01?

Solution 1

Because floating point arithmetic != real number arithmetic. An illustration of the difference due to imprecision is, for some floats a and b, (a+b)-b != a. This applies to any language using floats.

Since floating point are binary numbers with finite precision, there's a finite amount of representable numbers, which leads accuracy problems and surprises like this. Here's another interesting read: What Every Computer Scientist Should Know About Floating-Point Arithmetic.

---

Back to your problem, basically there is no way to accurately represent 34.99 or 0.01 in binary (just like in decimal, 1/3 = 0.3333...), so approximations are used instead. To get around the problem, you can:

1. Use round($result, 2) on the result to round it to 2 decimal places.

2. Use integers. If that's currency, say US dollars, then store $35.00 as 3500 and $34.99 as 3499, then divide the result by 100.

It's a pity that PHP doesn't have a decimal datatype like other languages do.

Solution 2

Floating point numbers, like all numbers, must be stored in memory as a string of 0's and 1's. It's all bits to the computer. How floating point differs from integer is in how we interpret the 0's and 1's when we want to look at them.

One bit is the "sign" (0 = positive, 1 = negative), 8 bits are the exponent (ranging from -128 to +127), 23 bits are the number known as the "mantissa" (fraction). So the binary representation of (S1)(P8)(M23) has the value (-1^S)M*2^P

The "mantissa" takes on a special form. In normal scientific notation we display the "one's place" along with the fraction. For instance:

4.39 x 10^2 = 439

In binary the "one's place" is a single bit. Since we ignore all the left-most 0's in scientific notation (we ignore any insignificant figures) the first bit is guaranteed to be a 1

1.101 x 2^3 = 1101 = 13

Since we are guaranteed that the first bit will be a 1, we remove this bit when storing the number to save space. So the above number is stored as just 101 (for the mantissa). The leading 1 is assumed

As an example, let's take the binary string

00000010010110000000000000000000

Breaking it into it's components:

Sign    Power           Mantissa
 0     00000100   10110000000000000000000
 +        +4             1.1011
 +        +4       1 + .5 + .125 + .0625
 +        +4             1.6875

Applying our simple formula:

(-1^S)M*2^P
(-1^0)(1.6875)*2^(+4)
(1)(1.6875)*(16)
27

In other words, 00000010010110000000000000000000 is 27 in floating point (according to IEEE-754 standards).

For many numbers there is no exact binary representation, however. Much like how 1/3 = 0.333.... repeating forever, 1/100 is 0.00000010100011110101110000..... with a repeating "10100011110101110000". A 32-bit computer can't store the entire number in floating point, however. So it makes its best guess.

0.0000001010001111010111000010100011110101110000

Sign    Power           Mantissa
 +        -7     1.01000111101011100001010
 0    -00000111   01000111101011100001010
 0     11111001   01000111101011100001010
01111100101000111101011100001010

(note that negative 7 is produced using 2's complement)

It should be immediately clear that 01111100101000111101011100001010 looks nothing like 0.01

More importantly, however, this contains a truncated version of a repeating decimal. The original decimal contained a repeating "10100011110101110000". We've simplified this to 01000111101011100001010

Translating this floating point number back into decimal via our formula we get 0.0099999979 (note that this is for a 32-bit computer. A 64-bit computer would have much more accuracy)

A Decimal Equivalent

If it helps to understand the problem better, let's look decimal scientific notation when dealing with repeating decimals.

Let's assume that we have 10 "boxes" to store digits. Therefore if we wanted to store a number like 1/16 we would write:

+---+---+---+---+---+---+---+---+---+---+
| + | 6 | . | 2 | 5 | 0 | 0 | e | - | 2 |
+---+---+---+---+---+---+---+---+---+---+

Which is clearly just 6.25 e -2, where e is shorthand for *10^(. We've allocated 4 boxes for the decimal even though we only needed 2 (padding with zeroes), and we've allocated 2 boxes for signs (one for the sign of the number, one of the sign of the exponent)

Using 10 boxes like this we can display numbers ranging from -9.9999 e -9 to +9.9999 e +9

This works fine for anything with 4 or fewer decimal places, but what happens when we try to store a number like 2/3?

+---+---+---+---+---+---+---+---+---+---+
| + | 6 | . | 6 | 6 | 6 | 7 | e | - | 1 |
+---+---+---+---+---+---+---+---+---+---+

This new number 0.66667 does not exactly equal 2/3. In fact, it's off by 0.000003333.... If we were to try and write 0.66667 in base 3, we would get 0.2000000000012... instead of 0.2

This problem may become more apparent if we take something with a larger repeating decimal, like 1/7. This has 6 repeating digits: 0.142857142857...

Storing this into our decimal computer we can only show 5 of these digits:

+---+---+---+---+---+---+---+---+---+---+
| + | 1 | . | 4 | 2 | 8 | 6 | e | - | 1 |
+---+---+---+---+---+---+---+---+---+---+

This number, 0.14286, is off by .000002857...

It's "close to correct", but it's not exactly correct, and so if we tried to write this number in base 7 we would get some hideous number instead of 0.1. In fact, plugging this into Wolfram Alpha we get: .10000022320335...

These minor fractional differences should look familiar to your 0.0099999979 (as opposed to 0.01)

Solution 3

There's plenty of answers here about why floating point numbers work the way they do...

But there's little talk of arbitrary precision (Pickle mentioned it). If you want (or need) exact precision, the only way to do it (for rational numbers at least) is to use the BC Math extension (which is really just a BigNum, Arbitrary Precision implementation...

To add two numbers:

$number = '12345678901234.1234567890';
$number2 = '1';
echo bcadd($number, $number2);

will result in 12345678901235.1234567890...

This is called arbitrary precision math. Basically all numbers are strings which are parsed for every operation and operations are performed on a digit by digit basis (think long division, but done by the library). So that means it's quite slow (in comparison to regular math constructs). But it's very powerful. You can multiply, add, subtract, divide, find modulo and exponentiate any number that has an exact string representation.

So you can't do 1/3 with 100% accuracy, since it has a repeating decimal (and hence isn't rational).

But, if you want to know what 1500.0015 squared is:

Using 32 bit floats (double precision) gives the estimated result of:

2250004.5000023

But bcmath gives the exact answer of:

2250004.50000225

It all depends on the precision you need.

Also, something else to note here. PHP can only represent either 32 bit or 64 bit integers (depending on your install). So if an integer exceeds the size of the native int type (2.1 billion for 32bit, 9.2 x10^18, or 9.2 billion billion for signed ints), PHP will convert the int into a float. While that's not immediately a problem (Since all ints smaller than the precision of the system's float are by definition directly representable as floats), if you try multiplying two together, it'll lose significant precision.

For example, given $n = '40000000002':

As a number, $n will be float(40000000002), which is fine since it's exactly represented. But if we square it, we get: float(1.60000000016E+21)

As a string (using BC math), $n will be exactly '40000000002'. And if we square it, we get: string(22) "1600000000160000000004"...

So if you need the precision with large numbers, or rational decimal points, you might want to look into bcmath...

Solution 4

bcadd() might be useful here.

<?PHP

$a = '35';
$b = '-34.99';

echo $a + $b;
echo '<br />';
echo bcadd($a,$b,2);

?>

(inefficient output for clarity)

First line gives me 0.009999999999998. Second gives me 0.01

Solution 5

Because 0.01 can't be represented exactly as sum of series of binary fractions. And that is how floats are stored in memory.

I guess it is not what you want to hear, but it is answer to question. For how to fix see other answers.

View more solutions

Comments

[Dennis Haarbrink]

-1 Because it is absolutely not an answer to the question.

[Andrey]

@Dennis Haarbrink well, you downvoted this, someone downvoted my answer. well, so what is answer then?

[Andrey]

i would add that 0.01 also can't be represented as is. this should be marked as correct, because it gives explanation and how to fix. but for increasing usefulness of it please explain a bit why fp != real, with all that binary stuff and precision lose

[Dennis Haarbrink]

@Andrey: Yeah, don't know why your answer got downvoted since it's pretty much the correct answer :) The best answer IMHO is by @ircmaxell in the comments on the OP.

[NullUserException]

That is not what the OP asked. PS: I didn't downvote you.

[Tomasz Kowalczyk]

@Dennis: edited my answer, please consider your vote again. ;]

[Andrey]

@NullUserException i think that your answer is better, but this one still fine, not for accept, but not for downvote. usually when people ask questions that way they usually want to know how to fix it, not the philosophy behind it.

[Tomasz Kowalczyk]

So thought I, but now both sides should be happy. ;] Still someone voted down. Argh ;]

[NullUserException]

@irc Thanks. I incorporated bits of your comment into the answer

[Dennis Haarbrink]

@Tomasz Kowalczyk: Well, you have received 3 up and 2 down votes, that totals 26rep. I figure that should be enough for your answer :)

[ircmaxell]

One pedantic note: There is a finite set of floats a and b where (a+b)-b == a. They simply need to have both a prime factor of 2, and be representable in the appropriate number of bits (about 7 decimal digits for single precision, 16 for double). So a = 0.5 and b = 0.25 works (and will always work for systems with 32 bit single precision floats). For floats that don't fit either or both of those preconditions, then (a+b)-b != a. But if both a and b fit those preconditions, then (a+b)-b == a should be true (but it's a finite set)...

[NullUserException]

@irc True; I used the wrong word there.

[Tomasz Kowalczyk]

It is not about the score, but an objective thoughts about an answer. I am really "fanatic" about programming and any disagreement must be clear whether I am right or someone is right. ;]

[stevendesu]

I'd give +1, but there are more links and less explanation than I'd like. Perhaps mention that the decimal value 0.01 in binary has a repeating "10100011110101110000" (the number looks like 0.00000010100011110101110000.....). Then further explain that a 32-bit computer is limited to expressing 23 significant digits (plus 8 for exponent and 1 for sign = 32 bits), meaning it becomes 0.00000010100011110101110000101 = d0.0099999979

[stevendesu]

Sum of series of binary what-now? That's not how floats are stored. A float is essentially scientific notation in binary. One bit is the "sign" (0 = positive, 1 = negative), 8 bits are the exponent (ranging from -128 to +127), 23 bits are the number known as the "mantissa". So the binary representation of (S1)(P8)(M23) has the value (-1^S)M*2^P

[NullUserException]

@steve Some of that is specific to 32-bit IEEE-754 floating point, but I am just mentioning floats in general

[Andrey]

@steven_desu thank you for lesson. key part here is that mantissa is stored as binary fraction. it is answer to question "why" can't decimal fractions be stored precisely.

[NikiC]

+1 thanks, now I know how floats are stored. PS: No, Windows doesn't. At least in PHP5.3.1/Win7 I did have floating point issues ;)

[Jrgns]

See @ircmaxell's answer below for more info on how to maintain accuracy when working with floats / decimal numbers.

[Mikko Rantalainen]

Nitpick: a number, such as 1/3, can have repeating decimal representation and still be rational. "Rational numbers" are all numbers that can be presented as a fraction of two numbers a and b where both a and b are integers. And 1/3 is indeed an example of such a number.

[mulllhausen]

+1 i came here looking for a method of dividing one huge string by another and found bcmath in your answer. thanks!

[Adam P. Goucher]

The last paragraph (which claims that the OS decides whether to round floating-point values) should be removed. The outcome of a floating-point calculation is mandated by IEEE 754, so "0.1 + 0.2 == 0.3" must evaluate to false on any compliant system. Some programs are dependent on floating-point operations behaving in this way.

[stevendesu]

@AdamP.Goucher I updated my post on February 15th per your comment. I neglected to comment here mentioning such, so I'm doing so now. Thanks for the improvement to the answer.

[stevendesu]

I think it's a bit close-minded to say the only way to do it is to use bc_math. I'd say the recommended way to do it is to you bc_math. You're free to implement your own system if you'd like :D It's just WAY more hassle than it's worth.