Consider the following code:
0.1 + 0.2 == 0.3 > false
0.1 + 0.2 > 0.30000000000000004
Why do these inaccuracies happen?
Binary floating point math is like this. In most programming languages, it is based on the IEEE 754 standard. The crux of the problem is that numbers are represented in this format as a whole number times a power of two; rational numbers (such as 0.1
, which is 1/10
) whose denominator is not a power of two cannot be exactly represented.
For 0.1
in the standard binary64
format, the representation can be written exactly as
0.1000000000000000055511151231257827021181583404541015625
in decimal, or0x1.999999999999ap4
in C99 hexfloat notation.
In contrast, the rational number 0.1
, which is 1/10
, can be written exactly as
0.1
in decimal, or0x1.99999999999999...p4
in an analogue of C99 hexfloat notation, where the...
represents an unending sequence of 9’s.
The constants 0.2
and 0.3
in your program will also be approximations to their true values. It happens that the closest double
to 0.2
is larger than the rational number 0.2
but that the closest double
to 0.3
is smaller than the rational number 0.3
. The sum of 0.1
and 0.2
winds up being larger than the rational number 0.3
and hence disagreeing with the constant in your code.
A fairly comprehensive treatment of floatingpoint arithmetic issues is What Every Computer Scientist Should Know About FloatingPoint Arithmetic. For an easiertodigest explanation, see floatingpointgui.de.
Side Note: All positional (baseN) number systems share this problem with precision
Plain old decimal (base 10) numbers have the same issues, which is why numbers like 1/3 end up as 0.333333333…
You’ve just stumbled on a number (3/10) that happens to be easy to represent with the decimal system, but doesn’t fit the binary system. It goes both ways (to some small degree) as well: 1/16 is an ugly number in decimal (0.0625), but in binary it looks as neat as a 10,000th does in decimal (0.0001)** – if we were in the habit of using a base2 number system in our daily lives, you’d even look at that number and instinctively understand you could arrive there by halving something, halving it again, and again and again.
** Of course, that’s not exactly how floatingpoint numbers are stored in memory (they use a form of scientific notation). However, it does illustrate the point that binary floatingpoint precision errors tend to crop up because the “real world” numbers we are usually interested in working with are so often powers of ten – but only because we use a decimal number system daytoday. This is also why we’ll say things like 71% instead of “5 out of every 7” (71% is an approximation, since 5/7 can’t be represented exactly with any decimal number).
So no: binary floating point numbers are not broken, they just happen to be as imperfect as every other baseN number system 🙂
Side Side Note: Working with Floats in Programming
In practice, this problem of precision means you need to use rounding functions to round your floating point numbers off to however many decimal places you’re interested in before you display them.
You also need to replace equality tests with comparisons that allow some amount of tolerance, which means:
Do not do if (x == y) { ... }
Instead do if (abs(x  y) < myToleranceValue) { ... }
.
where abs
is the absolute value. myToleranceValue
needs to be chosen for your particular application – and it will have a lot to do with how much “wiggle room” you are prepared to allow, and what the largest number you are going to be comparing may be (due to loss of precision issues). Beware of “epsilon” style constants in your language of choice. These are not to be used as tolerance values.
Answer：
A Hardware Designer’s Perspective
I believe I should add a hardware designer’s perspective to this since I design and build floating point hardware. Knowing the origin of the error may help in understanding what is happening in the software, and ultimately, I hope this helps explain the reasons for why floating point errors happen and seem to accumulate over time.
1. Overview
From an engineering perspective, most floating point operations will have some element of error since the hardware that does the floating point computations is only required to have an error of less than one half of one unit in the last place. Therefore, much hardware will stop at a precision that’s only necessary to yield an error of less than one half of one unit in the last place for a single operation which is especially problematic in floating point division. What constitutes a single operation depends upon how many operands the unit takes. For most, it is two, but some units take 3 or more operands. Because of this, there is no guarantee that repeated operations will result in a desirable error since the errors add up over time.
2. Standards
Most processors follow the IEEE754 standard but some use denormalized, or different standards
. For example, there is a denormalized mode in IEEE754 which allows representation of very small floating point numbers at the expense of precision. The following, however, will cover the normalized mode of IEEE754 which is the typical mode of operation.
In the IEEE754 standard, hardware designers are allowed any value of error/epsilon as long as it’s less than one half of one unit in the last place, and the result only has to be less than one half of one unit in the last place for one operation. This explains why when there are repeated operations, the errors add up. For IEEE754 double precision, this is the 54th bit, since 53 bits are used to represent the numeric part (normalized), also called the mantissa, of the floating point number (e.g. the 5.3 in 5.3e5). The next sections go into more detail on the causes of hardware error on various floating point operations.
3. Cause of Rounding Error in Division
The main cause of the error in floating point division is the division algorithms used to calculate the quotient. Most computer systems calculate division using multiplication by an inverse, mainly in Z=X/Y
, Z = X * (1/Y)
. A division is computed iteratively i.e. each cycle computes some bits of the quotient until the desired precision is reached, which for IEEE754 is anything with an error of less than one unit in the last place. The table of reciprocals of Y (1/Y) is known as the quotient selection table (QST) in the slow division, and the size in bits of the quotient selection table is usually the width of the radix, or a number of bits of the quotient computed in each iteration, plus a few guard bits. For the IEEE754 standard, double precision (64bit), it would be the size of the radix of the divider, plus a few guard bits k, where k>=2
. So for example, a typical Quotient Selection Table for a divider that computes 2 bits of the quotient at a time (radix 4) would be 2+2= 4
bits (plus a few optional bits).
3.1 Division Rounding Error: Approximation of Reciprocal
What reciprocals are in the quotient selection table depend on the division method: slow division such as SRT division, or fast division such as Goldschmidt division; each entry is modified according to the division algorithm in an attempt to yield the lowest possible error. In any case, though, all reciprocals are approximations of the actual reciprocal and introduce some element of error. Both slow division and fast division methods calculate the quotient iteratively, i.e. some number of bits of the quotient are calculated each step, then the result is subtracted from the dividend, and the divider repeats the steps until the error is less than one half of one unit in the last place. Slow division methods calculate a fixed number of digits of the quotient in each step and are usually less expensive to build, and fast division methods calculate a variable number of digits per step and are usually more expensive to build. The most important part of the division methods is that most of them rely upon repeated multiplication by an approximation of a reciprocal, so they are prone to error.
4. Rounding Errors in Other Operations: Truncation
Another cause of the rounding errors in all operations are the different modes of truncation of the final answer that IEEE754 allows. There’s truncate, roundtowardszero, roundtonearest (default), rounddown, and roundup. All methods introduce an element of error of less than one unit in the last place for a single operation. Over time and repeated operations, truncation also adds cumulatively to the resultant error. This truncation error is especially problematic in exponentiation, which involves some form of repeated multiplication.
5. Repeated Operations
Since the hardware that does the floating point calculations only needs to yield a result with an error of less than one half of one unit in the last place for a single operation, the error will grow over repeated operations if not watched. This is the reason that in computations that require a bounded error, mathematicians use methods such as using the roundtonearest even digit in the last place of IEEE754, because, over time, the errors are more likely to cancel each other out, and Interval Arithmetic combined with variations of the IEEE 754 rounding modes to predict rounding errors, and correct them. Because of its low relative error compared to other rounding modes, round to nearest even digit (in the last place), is the default rounding mode of IEEE754.
Note that the default rounding mode, roundtonearest even digit in the last place, guarantees an error of less than one half of one unit in the last place for one operation. Using the truncation, roundup, and round down alone may result in an error that is greater than one half of one unit in the last place, but less than one unit in the last place, so these modes are not recommended unless they are used in Interval Arithmetic.
6. Summary
In short, the fundamental reason for the errors in floating point operations is a combination of the truncation in hardware, and the truncation of a reciprocal in the case of division. Since the IEEE754 standard only requires an error of less than one half of one unit in the last place for a single operation, the floating point errors over repeated operations will add up unless corrected.
Answer：
When you convert .1 or 1/10 to base 2 (binary) you get a repeating pattern after the decimal point, just like trying to represent 1/3 in base 10. The value is not exact, and therefore you can’t do exact math with it using normal floating point methods.
Answer：
Most answers here address this question in very dry, technical terms. I’d like to address this in terms that normal human beings can understand.
Imagine that you are trying to slice up pizzas. You have a robotic pizza cutter that can cut pizza slices exactly in half. It can halve a whole pizza, or it can halve an existing slice, but in any case, the halving is always exact.
That pizza cutter has very fine movements, and if you start with a whole pizza, then halve that, and continue halving the smallest slice each time, you can do the halving 53 times before the slice is too small for even its highprecision abilities. At that point, you can no longer halve that very thin slice, but must either include or exclude it as is.
Now, how would you piece all the slices in such a way that would add up to onetenth (0.1) or onefifth (0.2) of a pizza? Really think about it, and try working it out. You can even try to use a real pizza, if you have a mythical precision pizza cutter at hand. 🙂
Most experienced programmers, of course, know the real answer, which is that there is no way to piece together an exact tenth or fifth of the pizza using those slices, no matter how finely you slice them. You can do a pretty good approximation, and if you add up the approximation of 0.1 with the approximation of 0.2, you get a pretty good approximation of 0.3, but it’s still just that, an approximation.
For doubleprecision numbers (which is the precision that allows you to halve your pizza 53 times), the numbers immediately less and greater than 0.1 are 0.09999999999999999167332731531132594682276248931884765625 and 0.1000000000000000055511151231257827021181583404541015625. The latter is quite a bit closer to 0.1 than the former, so a numeric parser will, given an input of 0.1, favour the latter.
(The difference between those two numbers is the “smallest slice” that we must decide to either include, which introduces an upward bias, or exclude, which introduces a downward bias. The technical term for that smallest slice is an ulp.)
In the case of 0.2, the numbers are all the same, just scaled up by a factor of 2. Again, we favour the value that’s slightly higher than 0.2.
Notice that in both cases, the approximations for 0.1 and 0.2 have a slight upward bias. If we add enough of these biases in, they will push the number further and further away from what we want, and in fact, in the case of 0.1 + 0.2, the bias is high enough that the resulting number is no longer the closest number to 0.3.
In particular, 0.1 + 0.2 is really 0.1000000000000000055511151231257827021181583404541015625 + 0.200000000000000011102230246251565404236316680908203125 = 0.3000000000000000444089209850062616169452667236328125, whereas the number closest to 0.3 is actually 0.299999999999999988897769753748434595763683319091796875.
P.S. Some programming languages also provide pizza cutters that can split slices into exact tenths. Although such pizza cutters are uncommon, if you do have access to one, you should use it when it’s important to be able to get exactly onetenth or onefifth of a slice.
Answer：
Floating point rounding errors. 0.1 cannot be represented as accurately in base2 as in base10 due to the missing prime factor of 5. Just as 1/3 takes an infinite number of digits to represent in decimal, but is “0.1” in base3, 0.1 takes an infinite number of digits in base2 where it does not in base10. And computers don’t have an infinite amount of memory.
Answer：
In addition to the other correct answers, you may want to consider scaling your values to avoid problems with floatingpoint arithmetic.
For example:
var result = 1.0 + 2.0; // result === 3.0 returns true
… instead of:
var result = 0.1 + 0.2; // result === 0.3 returns false
The expression 0.1 + 0.2 === 0.3
returns false
in JavaScript, but fortunately integer arithmetic in floatingpoint is exact, so decimal representation errors can be avoided by scaling.
As a practical example, to avoid floatingpoint problems where accuracy is paramount, it is recommended^{1} to handle money as an integer representing the number of cents: 2550
cents instead of 25.50
dollars.
^{1} Douglas Crockford: JavaScript: The Good Parts: Appendix A – Awful Parts (page 105).
Answer：
My answer is quite long, so I’ve split it into three sections. Since the question is about floating point mathematics, I’ve put the emphasis on what the machine actually does. I’ve also made it specific to double (64 bit) precision, but the argument applies equally to any floating point arithmetic.
Preamble
An IEEE 754 doubleprecision binary floatingpoint format (binary64) number represents a number of the form
value = (1)^s * (1.m_{51}m_{50}…m_{2}m_{1}m_{0})_{2} * 2^{e1023}
in 64 bits:
 The first bit is the sign bit:
1
if the number is negative,0
otherwise^{1}.  The next 11 bits are the exponent, which is offset by 1023. In other words, after reading the exponent bits from a doubleprecision number, 1023 must be subtracted to obtain the power of two.
 The remaining 52 bits are the significand (or mantissa). In the mantissa, an ‘implied’
1.
is always^{2} omitted since the most significant bit of any binary value is1
.
^{1} – IEEE 754 allows for the concept of a signed zero – +0
and 0
are treated differently: 1 / (+0)
is positive infinity; 1 / (0)
is negative infinity. For zero values, the mantissa and exponent bits are all zero. Note: zero values (+0 and 0) are explicitly not classed as denormal^{2}.
^{2} – This is not the case for denormal numbers, which have an offset exponent of zero (and an implied 0.
). The range of denormal double precision numbers is d_{min} ≤ x ≤ d_{max}, where d_{min} (the smallest representable nonzero number) is 2^{1023 – 51} (≈ 4.94 * 10^{324}) and d_{max} (the largest denormal number, for which the mantissa consists entirely of 1
s) is 2^{1023 + 1} – 2^{1023 – 51} (≈ 2.225 * 10^{308}).
Turning a double precision number to binary
Many online converters exist to convert a double precision floating point number to binary (e.g. at binaryconvert.com), but here is some sample C# code to obtain the IEEE 754 representation for a double precision number (I separate the three parts with colons (:
):
public static string BinaryRepresentation(double value)
{
long valueInLongType = BitConverter.DoubleToInt64Bits(value);
string bits = Convert.ToString(valueInLongType, 2);
string leadingZeros = new string('0', 64  bits.Length);
string binaryRepresentation = leadingZeros + bits;
string sign = binaryRepresentation[0].ToString();
string exponent = binaryRepresentation.Substring(1, 11);
string mantissa = binaryRepresentation.Substring(12);
return string.Format("{0}:{1}:{2}", sign, exponent, mantissa);
}
Getting to the point: the original question
(Skip to the bottom for the TL;DR version)
Cato Johnston (the question asker) asked why 0.1 + 0.2 != 0.3.
Written in binary (with colons separating the three parts), the IEEE 754 representations of the values are:
0.1 => 0:01111111011:1001100110011001100110011001100110011001100110011010
0.2 => 0:01111111100:1001100110011001100110011001100110011001100110011010
Note that the mantissa is composed of recurring digits of 0011
. This is key to why there is any error to the calculations – 0.1, 0.2 and 0.3 cannot be represented in binary precisely in a finite number of binary bits any more than 1/9, 1/3 or 1/7 can be represented precisely in decimal digits.
Also note that we can decrease the power in the exponent by 52 and shift the point in the binary representation to the right by 52 places (much like 10^{3} * 1.23 == 10^{5} * 123). This then enables us to represent the binary representation as the exact value that it represents in the form a * 2^{p}. where ‘a’ is an integer.
Converting the exponents to decimal, removing the offset, and readding the implied 1
(in square brackets), 0.1 and 0.2 are:
0.1 => 2^4 * [1].1001100110011001100110011001100110011001100110011010
0.2 => 2^3 * [1].1001100110011001100110011001100110011001100110011010
or
0.1 => 2^56 * 7205759403792794 = 0.1000000000000000055511151231257827021181583404541015625
0.2 => 2^55 * 7205759403792794 = 0.200000000000000011102230246251565404236316680908203125
To add two numbers, the exponent needs to be the same, i.e.:
0.1 => 2^3 * 0.1100110011001100110011001100110011001100110011001101(0)
0.2 => 2^3 * 1.1001100110011001100110011001100110011001100110011010
sum = 2^3 * 10.0110011001100110011001100110011001100110011001100111
or
0.1 => 2^55 * 3602879701896397 = 0.1000000000000000055511151231257827021181583404541015625
0.2 => 2^55 * 7205759403792794 = 0.200000000000000011102230246251565404236316680908203125
sum = 2^55 * 10808639105689191 = 0.3000000000000000166533453693773481063544750213623046875
Since the sum is not of the form 2^{n} * 1.{bbb} we increase the exponent by one and shift the decimal (binary) point to get:
sum = 2^2 * 1.0011001100110011001100110011001100110011001100110011(1)
= 2^54 * 5404319552844595.5 = 0.3000000000000000166533453693773481063544750213623046875
There are now 53 bits in the mantissa (the 53rd is in square brackets in the line above). The default rounding mode for IEEE 754 is ‘Round to Nearest‘ – i.e. if a number x falls between two values a and b, the value where the least significant bit is zero is chosen.
a = 2^54 * 5404319552844595 = 0.299999999999999988897769753748434595763683319091796875
= 2^2 * 1.0011001100110011001100110011001100110011001100110011
x = 2^2 * 1.0011001100110011001100110011001100110011001100110011(1)
b = 2^2 * 1.0011001100110011001100110011001100110011001100110100
= 2^54 * 5404319552844596 = 0.3000000000000000444089209850062616169452667236328125
Note that a and b differ only in the last bit; ...0011
+ 1
= ...0100
. In this case, the value with the least significant bit of zero is b, so the sum is:
sum = 2^2 * 1.0011001100110011001100110011001100110011001100110100
= 2^54 * 5404319552844596 = 0.3000000000000000444089209850062616169452667236328125
whereas the binary representation of 0.3 is:
0.3 => 2^2 * 1.0011001100110011001100110011001100110011001100110011
= 2^54 * 5404319552844595 = 0.299999999999999988897769753748434595763683319091796875
which only differs from the binary representation of the sum of 0.1 and 0.2 by 2^{54}.
The binary representation of 0.1 and 0.2 are the most accurate representations of the numbers allowable by IEEE 754. The addition of these representation, due to the default rounding mode, results in a value which differs only in the leastsignificantbit.
TL;DR
Writing 0.1 + 0.2
in a IEEE 754 binary representation (with colons separating the three parts) and comparing it to 0.3
, this is (I’ve put the distinct bits in square brackets):
0.1 + 0.2 => 0:01111111101:0011001100110011001100110011001100110011001100110[100]
0.3 => 0:01111111101:0011001100110011001100110011001100110011001100110[011]
Converted back to decimal, these values are:
0.1 + 0.2 => 0.300000000000000044408920985006...
0.3 => 0.299999999999999988897769753748...
The difference is exactly 2^{54}, which is ~5.5511151231258 × 10^{17} – insignificant (for many applications) when compared to the original values.
Comparing the last few bits of a floating point number is inherently dangerous, as anyone who reads the famous “What Every Computer Scientist Should Know About FloatingPoint Arithmetic” (which covers all the major parts of this answer) will know.
Most calculators use additional guard digits to get around this problem, which is how 0.1 + 0.2
would give 0.3
: the final few bits are rounded.
Answer：
Floating point numbers stored in the computer consist of two parts, an integer and an exponent that the base is taken to and multiplied by the integer part.
If the computer were working in base 10, 0.1
would be 1 x 10⁻¹
, 0.2
would be 2 x 10⁻¹
, and 0.3
would be 3 x 10⁻¹
. Integer math is easy and exact, so adding 0.1 + 0.2
will obviously result in 0.3
.
Computers don’t usually work in base 10, they work in base 2. You can still get exact results for some values, for example 0.5
is 1 x 2⁻¹
and 0.25
is 1 x 2⁻²
, and adding them results in 3 x 2⁻²
, or 0.75
. Exactly.
The problem comes with numbers that can be represented exactly in base 10, but not in base 2. Those numbers need to be rounded to their closest equivalent. Assuming the very common IEEE 64bit floating point format, the closest number to 0.1
is 3602879701896397 x 2⁻⁵⁵
, and the closest number to 0.2
is 7205759403792794 x 2⁻⁵⁵
; adding them together results in 10808639105689191 x 2⁻⁵⁵
, or an exact decimal value of 0.3000000000000000444089209850062616169452667236328125
. Floating point numbers are generally rounded for display.
Answer：
Floating point rounding error. From What Every Computer Scientist Should Know About FloatingPoint Arithmetic:
Squeezing infinitely many real numbers into a finite number of bits requires an approximate representation. Although there are infinitely many integers, in most programs the result of integer computations can be stored in 32 bits. In contrast, given any fixed number of bits, most calculations with real numbers will produce quantities that cannot be exactly represented using that many bits. Therefore the result of a floatingpoint calculation must often be rounded in order to fit back into its finite representation. This rounding error is the characteristic feature of floatingpoint computation.
Answer：
My workaround:
function add(a, b, precision) {
var x = Math.pow(10, precision  2);
return (Math.round(a * x) + Math.round(b * x)) / x;
}
precision refers to the number of digits you want to preserve after the decimal point during addition.
Answer：
A lot of good answers have been posted, but I’d like to append one more.
Not all numbers can be represented via floats/doubles
For example, the number “0.2” will be represented as “0.200000003” in single precision in IEEE754 float point standard.
Model for store real numbers under the hood represent float numbers as
Even though you can type 0.2
easily, FLT_RADIX
and DBL_RADIX
is 2; not 10 for a computer with FPU which uses “IEEE Standard for Binary FloatingPoint Arithmetic (ISO/IEEE Std 7541985)”.
So it is a bit hard to represent such numbers exactly. Even if you specify this variable explicitly without any intermediate calculation.
Answer：
Some statistics related to this famous double precision question.
When adding all values (a + b) using a step of 0.1 (from 0.1 to 100) we have ~15% chance of precision error. Note that the error could result in slightly bigger or smaller values.
Here are some examples:
0.1 + 0.2 = 0.30000000000000004 (BIGGER)
0.1 + 0.7 = 0.7999999999999999 (SMALLER)
...
1.7 + 1.9 = 3.5999999999999996 (SMALLER)
1.7 + 2.2 = 3.9000000000000004 (BIGGER)
...
3.2 + 3.6 = 6.800000000000001 (BIGGER)
3.2 + 4.4 = 7.6000000000000005 (BIGGER)
When subtracting all values (a – b where a > b) using a step of 0.1 (from 100 to 0.1) we have ~34% chance of precision error.
Here are some examples:
0.6  0.2 = 0.39999999999999997 (SMALLER)
0.5  0.4 = 0.09999999999999998 (SMALLER)
...
2.1  0.2 = 1.9000000000000001 (BIGGER)
2.0  1.9 = 0.10000000000000009 (BIGGER)
...
100  99.9 = 0.09999999999999432 (SMALLER)
100  99.8 = 0.20000000000000284 (BIGGER)
*15% and 34% are indeed huge, so always use BigDecimal when precision is of big importance. With 2 decimal digits (step 0.01) the situation worsens a bit more (18% and 36%).
Answer：
No, not broken, but most decimal fractions must be approximated
Summary
Floating point arithmetic is exact, unfortunately, it doesn’t match up well with our usual base10 number representation, so it turns out we are often giving it input that is slightly off from what we wrote.
Even simple numbers like 0.01, 0.02, 0.03, 0.04 … 0.24 are not representable exactly as binary fractions. If you count up 0.01, .02, .03 …, not until you get to 0.25 will you get the first fraction representable in base_{2}. If you tried that using FP, your 0.01 would have been slightly off, so the only way to add 25 of them up to a nice exact 0.25 would have required a long chain of causality involving guard bits and rounding. It’s hard to predict so we throw up our hands and say “FP is inexact”, but that’s not really true.
We constantly give the FP hardware something that seems simple in base 10 but is a repeating fraction in base 2.
How did this happen?
When we write in decimal, every fraction (specifically, every terminating decimal) is a rational number of the form
a / (2^{n} x 5^{m})
In binary, we only get the 2^{n} term, that is:
a / 2^{n}
So in decimal, we can’t represent ^{1}/_{3}. Because base 10 includes 2 as a prime factor, every number we can write as a binary fraction also can be written as a base 10 fraction. However, hardly anything we write as a base_{10} fraction is representable in binary. In the range from 0.01, 0.02, 0.03 … 0.99, only three numbers can be represented in our FP format: 0.25, 0.50, and 0.75, because they are 1/4, 1/2, and 3/4, all numbers with a prime factor using only the 2^{n} term.
In base_{10} we can’t represent ^{1}/_{3}. But in binary, we can’t do ^{1}/_{10} or ^{1}/_{3}.
So while every binary fraction can be written in decimal, the reverse is not true. And in fact most decimal fractions repeat in binary.
Dealing with it
Developers are usually instructed to do < epsilon comparisons, better advice might be to round to integral values (in the C library: round() and roundf(), i.e., stay in the FP format) and then compare. Rounding to a specific decimal fraction length solves most problems with output.
Also, on real numbercrunching problems (the problems that FP was invented for on early, frightfully expensive computers) the physical constants of the universe and all other measurements are only known to a relatively small number of significant figures, so the entire problem space was “inexact” anyway. FP “accuracy” isn’t a problem in this kind of application.
The whole issue really arises when people try to use FP for bean counting. It does work for that, but only if you stick to integral values, which kind of defeats the point of using it. This is why we have all those decimal fraction software libraries.
I love the Pizza answer by Chris, because it describes the actual problem, not just the usual handwaving about “inaccuracy”. If FP were simply “inaccurate”, we could fix that and would have done it decades ago. The reason we haven’t is because the FP format is compact and fast and it’s the best way to crunch a lot of numbers. Also, it’s a legacy from the space age and arms race and early attempts to solve big problems with very slow computers using small memory systems. (Sometimes, individual magnetic cores for 1bit storage, but that’s another story.)
Conclusion
If you are just counting beans at a bank, software solutions that use decimal string representations in the first place work perfectly well. But you can’t do quantum chromodynamics or aerodynamics that way.
Answer：
Did you try the duct tape solution?
Try to determine when errors occur and fix them with short if statements, it’s not pretty but for some problems it is the only solution and this is one of them.
if( (n * 0.1) < 100.0 ) { return n * 0.1  0.000000000000001 ;}
else { return n * 0.1 + 0.000000000000001 ;}
I had the same problem in a scientific simulation project in c#, and I can tell you that if you ignore the butterfly effect it’s gonna turn to a big fat dragon and bite you in the a**
Answer：
In order to offer The best solution I can say I discovered following method:
parseFloat((0.1 + 0.2).toFixed(10)) => Will return 0.3
Let me explain why it’s the best solution.
As others mentioned in above answers it’s a good idea to use ready to use Javascript toFixed() function to solve the problem. But most likely you’ll encounter with some problems.
Imagine you are going to add up two float numbers like 0.2
and 0.7
here it is: 0.2 + 0.7 = 0.8999999999999999
.
Your expected result was 0.9
it means you need a result with 1 digit precision in this case.
So you should have used (0.2 + 0.7).tofixed(1)
but you can’t just give a certain parameter to toFixed() since it depends on the given number, for instance
`0.22 + 0.7 = 0.9199999999999999`
In this example you need 2 digits precision so it should be toFixed(2)
, so what should be the paramter to fit every given float number?
You might say let it be 10 in every situation then:
(0.2 + 0.7).toFixed(10) => Result will be "0.9000000000"
Damn! What are you going to do with those unwanted zeros after 9?
It’s the time to convert it to float to make it as you desire:
parseFloat((0.2 + 0.7).toFixed(10)) => Result will be 0.9
Now that you found the solution, it’s better to offer it as a function like this:
function floatify(number){
return parseFloat((number).toFixed(10));
}
Let’s try it yourself:
function floatify(number){
return parseFloat((number).toFixed(10));
}
function addUp(){
var number1 = +$("#number1").val();
var number2 = +$("#number2").val();
var unexpectedResult = number1 + number2;
var expectedResult = floatify(number1 + number2);
$("#unexpectedResult").text(unexpectedResult);
$("#expectedResult").text(expectedResult);
}
addUp();
input{
width: 50px;
}
#expectedResult{
color: green;
}
#unexpectedResult{
color: red;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="number1" value="0.2" onclick="addUp()" onkeyup="addUp()"/> +
<input id="number2" value="0.7" onclick="addUp()" onkeyup="addUp()"/> =
<p>Expected Result: <span id="expectedResult"></span></p>
<p>Unexpected Result: <span id="unexpectedResult"></span></p>
You can use it this way:
var x = 0.2 + 0.7;
floatify(x); => Result: 0.9
As W3SCHOOLS suggests there is another solution too, you can multiply and divide to solve the problem above:
var x = (0.2 * 10 + 0.1 * 10) / 10; // x will be 0.3
Keep in mind that (0.2 + 0.1) * 10 / 10
won’t work at all although it seems the same!
I prefer the first solution since I can apply it as a function which converts the input float to accurate output float.
Answer：
Those weird numbers appear because computers use binary(base 2) number system for calculation purposes, while we use decimal(base 10).
There are a majority of fractional numbers that cannot be represented precisely either in binary or in decimal or both. Result – A rounded up (but precise) number results.
Answer：
Many of this question’s numerous duplicates ask about the effects of floating point rounding on specific numbers. In practice, it is easier to get a feeling for how it works by looking at exact results of calculations of interest rather than by just reading about it. Some languages provide ways of doing that – such as converting a float
or double
to BigDecimal
in Java.
Since this is a languageagnostic question, it needs languageagnostic tools, such as a Decimal to FloatingPoint Converter.
Applying it to the numbers in the question, treated as doubles:
0.1 converts to 0.1000000000000000055511151231257827021181583404541015625,
0.2 converts to 0.200000000000000011102230246251565404236316680908203125,
0.3 converts to 0.299999999999999988897769753748434595763683319091796875, and
0.30000000000000004 converts to 0.3000000000000000444089209850062616169452667236328125.
Adding the first two numbers manually or in a decimal calculator such as Full Precision Calculator, shows the exact sum of the actual inputs is 0.3000000000000000166533453693773481063544750213623046875.
If it were rounded down to the equivalent of 0.3 the rounding error would be 0.0000000000000000277555756156289135105907917022705078125. Rounding up to the equivalent of 0.30000000000000004 also gives rounding error 0.0000000000000000277555756156289135105907917022705078125. The roundtoeven tie breaker applies.
Returning to the floating point converter, the raw hexadecimal for 0.30000000000000004 is 3fd3333333333334, which ends in an even digit and therefore is the correct result.
Answer：
Given that nobody has mentioned this…
Some high level languages such as Python and Java come with tools to overcome binary floating point limitations. For example:

Python’s
decimal
module and Java’sBigDecimal
class, that represent numbers internally with decimal notation (as opposed to binary notation). Both have limited precision, so they are still error prone, however they solve most common problems with binary floating point arithmetic.Decimals are very nice when dealing with money: ten cents plus twenty cents are always exactly thirty cents:
>>> 0.1 + 0.2 == 0.3 False >>> Decimal('0.1') + Decimal('0.2') == Decimal('0.3') True
Python’s
decimal
module is based on IEEE standard 8541987. 
Python’s
fractions
module and Apache Common’sBigFraction
class. Both represent rational numbers as(numerator, denominator)
pairs and they may give more accurate results than decimal floating point arithmetic.
Neither of these solutions is perfect (especially if we look at performances, or if we require a very high precision), but still they solve a great number of problems with binary floating point arithmetic.
Answer：
Can I just add; people always assume this to be a computer problem, but if you count with your hands (base 10), you can’t get (1/3+1/3=2/3)=true
unless you have infinity to add 0.333… to 0.333… so just as with the (1/10+2/10)!==3/10
problem in base 2, you truncate it to 0.333 + 0.333 = 0.666 and probably round it to 0.667 which would be also be technically inaccurate.
Count in ternary, and thirds are not a problem though – maybe some race with 15 fingers on each hand would ask why your decimal math was broken…
Answer：
The kind of floatingpoint math that can be implemented in a digital computer necessarily uses an approximation of the real numbers and operations on them. (The standard version runs to over fifty pages of documentation and has a committee to deal with its errata and further refinement.)
This approximation is a mixture of approximations of different kinds, each of which can either be ignored or carefully accounted for due to its specific manner of deviation from exactitude. It also involves a number of explicit exceptional cases at both the hardware and software levels that most people walk right past while pretending not to notice.
If you need infinite precision (using the number π, for example, instead of one of its many shorter standins), you should write or use a symbolic math program instead.
But if you’re okay with the idea that sometimes floatingpoint math is fuzzy in value and logic and errors can accumulate quickly, and you can write your requirements and tests to allow for that, then your code can frequently get by with what’s in your FPU.
Answer：
Just for fun, I played with the representation of floats, following the definitions from the Standard C99 and I wrote the code below.
The code prints the binary representation of floats in 3 separated groups
SIGN EXPONENT FRACTION
and after that it prints a sum, that, when summed with enough precision, it will show the value that really exists in hardware.
So when you write float x = 999...
, the compiler will transform that number in a bit representation printed by the function xx
such that the sum printed by the function yy
be equal to the given number.
In reality, this sum is only an approximation. For the number 999,999,999 the compiler will insert in bit representation of the float the number 1,000,000,000
After the code I attach a console session, in which I compute the sum of terms for both constants (minus PI and 999999999) that really exists in hardware, inserted there by the compiler.
#include <stdio.h>
#include <limits.h>
void
xx(float *x)
{
unsigned char i = sizeof(*x)*CHAR_BIT1;
do {
switch (i) {
case 31:
printf("sign:");
break;
case 30:
printf("exponent:");
break;
case 23:
printf("fraction:");
break;
}
char b=(*(unsigned long long*)x&((unsigned long long)1<<i))!=0;
printf("%d ", b);
} while (i);
printf("\n");
}
void
yy(float a)
{
int sign=!(*(unsigned long long*)&a&((unsigned long long)1<<31));
int fraction = ((1<<23)1)&(*(int*)&a);
int exponent = (255&((*(int*)&a)>>23))127;
printf(sign?"positive" " ( 1+":"negative" " ( 1+");
unsigned int i = 1<<22;
unsigned int j = 1;
do {
char b=(fraction&i)!=0;
b&&(printf("1/(%d) %c", 1<<j, (fraction&(i1))?'+':')' ), 0);
} while (j++, i>>=1);
printf("*2^%d", exponent);
printf("\n");
}
void
main()
{
float x=3.14;
float y=999999999;
printf("%lu\n", sizeof(x));
xx(&x);
xx(&y);
yy(x);
yy(y);
}
Here is a console session in which I compute the real value of the float that exists in hardware. I used bc
to print the sum of terms outputted by the main program. One can insert that sum in python repl
or something similar also.
 .../terra1/stub
@ qemacs f.c
 .../terra1/stub
@ gcc f.c
 .../terra1/stub
@ ./a.out
sign:1 exponent:1 0 0 0 0 0 0 fraction:0 1 0 0 1 0 0 0 1 1 1 1 0 1 0 1 1 1 0 0 0 0 1 1
sign:0 exponent:1 0 0 1 1 1 0 fraction:0 1 1 0 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 0
negative ( 1+1/(2) +1/(16) +1/(256) +1/(512) +1/(1024) +1/(2048) +1/(8192) +1/(32768) +1/(65536) +1/(131072) +1/(4194304) +1/(8388608) )*2^1
positive ( 1+1/(2) +1/(4) +1/(16) +1/(32) +1/(64) +1/(512) +1/(1024) +1/(4096) +1/(16384) +1/(32768) +1/(262144) +1/(1048576) )*2^29
 .../terra1/stub
@ bc
scale=15
( 1+1/(2) +1/(4) +1/(16) +1/(32) +1/(64) +1/(512) +1/(1024) +1/(4096) +1/(16384) +1/(32768) +1/(262144) +1/(1048576) )*2^29
999999999.999999446351872
That’s it. The value of 999999999 is in fact
999999999.999999446351872
You can also check with bc
that 3.14 is also perturbed. Do not forget to set a scale
factor in bc
.
The displayed sum is what inside the hardware. The value you obtain by computing it depends on the scale you set. I did set the scale
factor to 15. Mathematically, with infinite precision, it seems it is 1,000,000,000.
Answer：
Another way to look at this: Used are 64 bits to represent numbers. As consequence there is no way more than 2**64 = 18,446,744,073,709,551,616 different numbers can be precisely represented.
However, Math says there are already infinitely many decimals between 0 and 1. IEE 754 defines an encoding to use these 64 bits efficiently for a much larger number space plus NaN and +/ Infinity, so there are gaps between accurately represented numbers filled with numbers only approximated.
Unfortunately 0.3 sits in a gap.
Answer：
Since this thread branched off a bit into a general discussion over current floating point implementations I’d add that there are projects on fixing their issues.
Take a look at https://posithub.org/ for example, which showcases a number type called posit (and its predecessor unum) that promises to offer better accuracy with fewer bits. If my understanding is correct, it also fixes the kind of problems in the question. Quite interesting project, the person behind it is a mathematician it Dr. John Gustafson. The whole thing is open source, with many actual implementations in C/C++, Python, Julia and C# (https://hastlayer.com/arithmetics).
Answer：
Imagine working in base ten with, say, 8 digits of accuracy. You check whether
1/3 + 2 / 3 == 1
and learn that this returns false
. Why? Well, as real numbers we have
1/3 = 0.333…. and 2/3 = 0.666….
Truncating at eight decimal places, we get
0.33333333 + 0.66666666 = 0.99999999
which is, of course, different from 1.00000000
by exactly 0.00000001
.
The situation for binary numbers with a fixed number of bits is exactly analogous. As real numbers, we have
1/10 = 0.0001100110011001100… (base 2)
and
1/5 = 0.0011001100110011001… (base 2)
If we truncated these to, say, seven bits, then we’d get
0.0001100 + 0.0011001 = 0.0100101
while on the other hand,
3/10 = 0.01001100110011… (base 2)
which, truncated to seven bits, is 0.0100110
, and these differ by exactly 0.0000001
.
The exact situation is slightly more subtle because these numbers are typically stored in scientific notation. So, for instance, instead of storing 1/10 as 0.0001100
we may store it as something like 1.10011 * 2^4
, depending on how many bits we’ve allocated for the exponent and the mantissa. This affects how many digits of precision you get for your calculations.
The upshot is that because of these rounding errors you essentially never want to use == on floatingpoint numbers. Instead, you can check if the absolute value of their difference is smaller than some fixed small number.
Answer：
It’s actually pretty simple. When you have a base 10 system (like ours), it can only express fractions that use a prime factor of the base. The prime factors of 10 are 2 and 5. So 1/2, 1/4, 1/5, 1/8, and 1/10 can all be expressed cleanly because the denominators all use prime factors of 10. In contrast, 1/3, 1/6, and 1/7 are all repeating decimals because their denominators use a prime factor of 3 or 7. In binary (or base 2), the only prime factor is 2. So you can only express fractions cleanly which only contain 2 as a prime factor. In binary, 1/2, 1/4, 1/8 would all be expressed cleanly as decimals. While, 1/5 or 1/10 would be repeating decimals. So 0.1 and 0.2 (1/10 and 1/5) while clean decimals in a base 10 system, are repeating decimals in the base 2 system the computer is operating in. When you do math on these repeating decimals, you end up with leftovers which carry over when you convert the computer’s base 2 (binary) number into a more human readable base 10 number.
Answer：
Since Python 3.5 you can use math.isclose()
function for testing approximate equality:
>>> import math
>>> math.isclose(0.1 + 0.2, 0.3)
True
>>> 0.1 + 0.2 == 0.3
False
Answer：
Decimal numbers such as 0.1
, 0.2
, and 0.3
are not represented exactly in binary encoded floating point types. The sum of the approximations for 0.1
and 0.2
differs from the approximation used for 0.3
, hence the falsehood of 0.1 + 0.2 == 0.3
as can be seen more clearly here:
#include <stdio.h>
int main() {
printf("0.1 + 0.2 == 0.3 is %s\n", 0.1 + 0.2 == 0.3 ? "true" : "false");
printf("0.1 is %.23f\n", 0.1);
printf("0.2 is %.23f\n", 0.2);
printf("0.1 + 0.2 is %.23f\n", 0.1 + 0.2);
printf("0.3 is %.23f\n", 0.3);
printf("0.3  (0.1 + 0.2) is %g\n", 0.3  (0.1 + 0.2));
return 0;
}
Output:
0.1 + 0.2 == 0.3 is false
0.1 is 0.10000000000000000555112
0.2 is 0.20000000000000001110223
0.1 + 0.2 is 0.30000000000000004440892
0.3 is 0.29999999999999998889777
0.3  (0.1 + 0.2) is 5.55112e17
For these computations to be evaluated more reliably, you would need to use a decimalbased representation for floating point values. The C Standard does not specify such types by default but as an extension described in a technical Report.
The _Decimal32
, _Decimal64
and _Decimal128
types might be available on your system (for example, GCC supports them on selected targets, but Clang does not support them on OS X).
Answer：
Math.sum ( javascript ) …. kind of operator replacement
.1 + .0001 + .1 > 0.00010000000000000286
Math.sum(.1 , .0001, .1) > 0.0001
Object.defineProperties(Math, {
sign: {
value: function (x) {
return x ? x < 0 ? 1 : 1 : 0;
}
},
precision: {
value: function (value, precision, type) {
var v = parseFloat(value),
p = Math.max(precision, 0)  0,
t = type  'round';
return (Math[t](v * Math.pow(10, p)) / Math.pow(10, p)).toFixed(p);
}
},
scientific_to_num: { // this is from https://gist.github.com/jiggzson
value: function (num) {
//if the number is in scientific notation remove it
if (/e/i.test(num)) {
var zero = '0',
parts = String(num).toLowerCase().split('e'), //split into coeff and exponent
e = parts.pop(), //store the exponential part
l = Math.abs(e), //get the number of zeros
sign = e / l,
coeff_array = parts[0].split('.');
if (sign === 1) {
num = zero + '.' + new Array(l).join(zero) + coeff_array.join('');
} else {
var dec = coeff_array[1];
if (dec)
l = l  dec.length;
num = coeff_array.join('') + new Array(l + 1).join(zero);
}
}
return num;
}
}
get_precision: {
value: function (number) {
var arr = Math.scientific_to_num((number + "")).split(".");
return arr[1] ? arr[1].length : 0;
}
},
diff:{
value: function(A,B){
var prec = this.max(this.get_precision(A),this.get_precision(B));
return +this.precision(AB,prec);
}
},
sum: {
value: function () {
var prec = 0, sum = 0;
for (var i = 0; i < arguments.length; i++) {
prec = this.max(prec, this.get_precision(arguments[i]));
sum += +arguments[i]; // force float to convert strings to number
}
return Math.precision(sum, prec);
}
}
});
the idea is to use Math instead operators to avoid float errors
Math.diff(0.2, 0.11) == 0.09 // true
0.2  0.11 == 0.09 // false
also note that Math.diff and Math.sum autodetect the precision to use
Math.sum accepts any number of arguments
Answer：
I just saw this interesting issue around floating points:
Consider the following results:
error = (2**53+1)  int(float(2**53+1))
>>> (2**53+1)  int(float(2**53+1))
1
We can clearly see a breakpoint when 2**53+1
– all works fine until 2**53
.
>>> (2**53)  int(float(2**53))
0
This happens because of the doubleprecision binary: IEEE 754 doubleprecision binary floatingpoint format: binary64
From the Wikipedia page for Doubleprecision floatingpoint format:
Doubleprecision binary floatingpoint is a commonly used format on PCs, due to its wider range over singleprecision floating point, in spite of its performance and bandwidth cost. As with singleprecision floatingpoint format, it lacks precision on integer numbers when compared with an integer format of the same size. It is commonly known simply as double. The IEEE 754 standard specifies a binary64 as having:
 Sign bit: 1 bit
 Exponent: 11 bits
 Significant precision: 53 bits (52 explicitly stored)
The real value assumed by a given 64bit doubleprecision datum with a given biased exponent and a 52bit fraction is
or
Thanks to @a_guest for pointing that out to me.
Answer：
A different question has been named as a duplicate to this one:
In C++, why is the result of cout << x
different from the value that a debugger is showing for x
?
The x
in the question is a float
variable.
One example would be
float x = 9.9F;
The debugger shows 9.89999962
, the output of cout
operation is 9.9
.
The answer turns out to be that cout
‘s default precision for float
is 6, so it rounds to 6 decimal digits.
See here for reference
Tags: math, mathexception