How to Deal With Floating Point Number Precision in JavaScript

How to deal with floating point number precision in JavaScript?

From the Floating-Point Guide:

What can I do to avoid this problem?

That depends on what kind of
calculations you’re doing.

• If you really need your results to add up exactly, especially when you
  work with money: use a special decimal
  datatype.
• If you just don’t want to see all those extra decimal places: simply
  format your result rounded to a fixed
  number of decimal places when
  displaying it.
• If you have no decimal datatype available, an alternative is to work
  with integers, e.g. do money
  calculations entirely in cents. But
  this is more work and has some
  drawbacks.

Note that the first point only applies if you really need specific precise decimal behaviour. Most people don't need that, they're just irritated that their programs don't work correctly with numbers like 1/10 without realizing that they wouldn't even blink at the same error if it occurred with 1/3.

If the first point really applies to you, use BigDecimal for JavaScript, which is not elegant at all, but actually solves the problem rather than providing an imperfect workaround.

Algorithm to correct the floating point precision error in JavaScript

When a decimal numeral is rounded to binary floating-point, there is no way to know, just from the result, what the original number was or how many significant digits it had. Infinitely many decimal numerals will round to the same result.

However, the rounding error is bounded. If it is known that the original number had at most a certain number of digits, then only decimal numerals with that number of digits are candidates. If only one of those candidates differs from the binary value by less than the maximum rounding error, then that one must be the original number.

If I recall correctly (I do not use JavaScript regularly), JavaScript uses IEEE-754 64-bit binary. For this format, it is known that any 15-digit decimal numeral may be converted to this binary floating-point format and back without error. Thus, if the original input was a decimal numeral with at most 15 significant digits, and it was converted to 64-bit binary floating-point (and no other operations were performed on it that could have introduced additional error), and you format the binary floating-point value as a 15-digit decimal numeral, you will have the original number.

The resulting decimal numeral may have trailing zeroes. It is not possible to know (from the binary floating-point value alone) whether those were in the original numeral.

How to deal with floating point number precision in JavaScript?

From the Floating-Point Guide:

What can I do to avoid this problem?

That depends on what kind of
calculations you’re doing.

• If you really need your results to add up exactly, especially when you
  work with money: use a special decimal
  datatype.
• If you just don’t want to see all those extra decimal places: simply
  format your result rounded to a fixed
  number of decimal places when
  displaying it.
• If you have no decimal datatype available, an alternative is to work
  with integers, e.g. do money
  calculations entirely in cents. But
  this is more work and has some
  drawbacks.

Note that the first point only applies if you really need specific precise decimal behaviour. Most people don't need that, they're just irritated that their programs don't work correctly with numbers like 1/10 without realizing that they wouldn't even blink at the same error if it occurred with 1/3.

If the first point really applies to you, use BigDecimal for JavaScript, which is not elegant at all, but actually solves the problem rather than providing an imperfect workaround.

Dealing with float precision in Javascript

From this post: How to deal with floating point number precision in JavaScript?

You have a few options:

• Use a special datatype for decimals, like decimal.js
• Format your result to some fixed number of significant digits, like this:
  (Math.floor(y/x) * x).toFixed(2)
• Convert all your numbers to integers

Is floating point math broken?

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, or
• 0x1.999999999999ap-4 in C99 hexfloat notation.

In contrast, the rational number 0.1, which is 1/10, can be written exactly as

• 0.1 in decimal, or
• 0x1.99999999999999...p-4 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 floating-point arithmetic issues is What Every Computer Scientist Should Know About Floating-Point Arithmetic. For an easier-to-digest explanation, see floating-point-gui.de.

Side Note: All positional (base-N) 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 base-2 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 floating-point numbers are stored in memory (they use a form of scientific notation). However, it does illustrate the point that binary floating-point 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 day-to-day. 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 base-N 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.

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