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.

## Best Solution

The SSE ops will subtly change the behavior of this algorithm because they don't have the intermediate 80-bit representation -- the math truly is done in 32 or 64 bits. The good news is that you can easily test it and see if it changes your results by simply specifying the /ARCH:SSE2 command line option to MSVC, which will cause it to use the SSE scalar ops instead of x87 FPU instructions for ordinary floating point math.

I'm not sure offhand of what the exact rounding behavior is around the integer boundaries, but you can test to see what'll happen when 1.999.. gets rounded from 64 to 32 bits by

egEdit, result:original poster ran this test and found that with truncation, the 1.99999 will round up to 2 both with and without /arch:SSE2.