When I do the following double multiplication in C# 100.0 * 1.005 I get 100,49999999999999 as a result. I believe this is because the exact number (or some intermedia result when evaluting the expression) cannot be represented. When I do the same computation in calc.exe I get 100.5 as expected.
Another example is the ninefold incrementation of 0.001 (that is the first time a deviation occurs) so basically 9d * 0.001d = 0,0090000000000000011. When I do the same computation in calc.exe I get 0.009 as expected.
Now one can argue, that I should choose decimal instead. But with decimal I get the problem with other computations for example with ((1M / 3M) * 3M) = 0,9999999999999999999999999999 while calc.exe says 1.
With calc.exe I can divide 1 by 3 several times until some real small number and then multiply with 3 again as many several times and then I reach exacty 1. I therefore suspect, that calc.exe computes internally with fractions, but obviously with real big ones, because it computes
(677605234775492641 / 116759166847407000) + (932737194383944703 / 2451942503795547000)
where the common denominator is -3422539506717149376 (an overflow occured) when doing a long computation, so it must be at least ulong. Does anybody know how computation in calc.exe is implemented? Is this implementation made somewhere public for reuse?
As described here, calc uses an arbitrary-precision engine for its calculations, while double is standard IEEE-754 arithmetic, and decimal is also floating-point arithmetic, just in decimal, which, as you point out, has the same problems, just in another base.
You can try finding such an arbitrary-precision arithmetic library for C# and use it, e.g. this one (no idea whether it's good; was the first result). The one inside calc is not available as an API, so you cannot use it.
Another point is that when you round the result to a certain number of places (less than 15), you'd also get the intuitively "correct" result in a lot of cases. C# already does some rounding to hide the exact value of a double from you (where 0.3 is definitely not exactly 0.3, but probably something like 0.30000000000000004). By reducing the number of digits you display you lessen the occurrence of such very small differences from the correct value.
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Is floating point math broken?
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If I execute the following expression in C#:
double i = 10*0.69;
i is: 6.8999999999999995. Why?
I understand numbers such as 1/3 can be hard to represent in binary as it has infinite recurring decimal places but this is not the case for 0.69. And 0.69 can easily be represented in binary, one binary number for 69 and another to denote the position of the decimal place.
How do I work around this? Use the decimal type?
Because you've misunderstood floating point arithmetic and how data is stored.
In fact, your code isn't actually performing any arithmetic at execution time in this particular case - the compiler will have done it, then saved a constant in the generated executable. However, it can't store an exact value of 6.9, because that value cannot be precisely represented in floating point point format, just like 1/3 can't be precisely stored in a finite decimal representation.
See if this article helps you.
why doesn't the framework work around this and hide this problem from me and give me the
right answer,0.69!!!
Stop behaving like a dilbert manager, and accept that computers, though cool and awesome, have limits. In your specific case, it doesn't just "hide" the problem, because you have specifically told it not to. The language (the computer) provides alternatives to the format, that you didn't choose. You chose double, which has certain advantages over decimal, and certain downsides. Now, knowing the answer, you're upset that the downsides don't magically disappear.
As a programmer, you are responsible for hiding this downside from managers, and there are many ways to do that. However, the makers of C# have a responsibility to make floating point work correctly, and correct floating point will occasionally result in incorrect math.
So will every other number storage method, as we do not have infinite bits. Our job as programmers is to work with limited resources to make cool things happen. They got you 90% of the way there, just get the torch home.
And 0.69 can easily be represented in
binary, one binary number for 69 and
another to denote the position of the
decimal place.
I think this is a common mistake - you're thinking of floating point numbers as if they are base-10 (i.e decimal - hence my emphasis).
So - you're thinking that there are two whole-number parts to this double: 69 and divide by 100 to get the decimal place to move - which could also be expressed as:
69 x 10 to the power of -2.
However floats store the 'position of the point' as base-2.
Your float actually gets stored as:
68999999999999995 x 2 to the power of some big negative number
This isn't as much of a problem once you're used to it - most people know and expect that 1/3 can't be expressed accurately as a decimal or percentage. It's just that the fractions that can't be expressed in base-2 are different.
but why doesn't the framework work around this and hide this problem from me and give me the right answer,0.69!!!
Because you told it to use binary floating point, and the solution is to use decimal floating point, so you are suggesting that the framework should disregard the type you specified and use decimal instead, which is very much slower because it is not directly implemented in hardware.
A more efficient solution is to not output the full value of the representation and explicitly specify the accuracy required by your output. If you format the output to two decimal places, you will see the result you expect. However if this is a financial application decimal is precisely what you should use - you've seen Superman III (and Office Space) haven't you ;)
Note that it is all a finite approximation of an infinite range, it is merely that decimal and double use a different set of approximations. The advantage of decimal is it produces the same approximations that you would if you were performing the calculation yourself. For example if you calculated 1/3, you would eventually stop writing 3's when it was 'good enough'.
For the same reason that 1 / 3 in a decimal systems comes out as 0.3333333333333333333333333333333333333333333 and not the exact fraction, which is infinitely long.
To work around it (e.g. to display on screen) try this:
double i = (double) Decimal.Multiply(10, (Decimal) 0.69);
Everyone seems to have answered your first question, but ignored the second part.
I know this has been discussed time and time again, but I can't seem to get even the most simple example of a one-step division of doubles to result in the expected, unrounded outcome in C# - so I'm wondering if perhaps there's i.e. some compiler flag or something else strange I'm not thinking of. Consider this example:
double v1 = 0.7;
double v2 = 0.025;
double result = v1 / v2;
When I break after the last line and examine it in the VS debugger, the value of "result" is 27.999999999999996. I'm aware that I can resolve it by changing to "decimal," but that's not possible in the case of the surrounding program. Is it not strange that two low-precision doubles like this can't divide to the correct value of 28? Is the only solution really to Math.Round the result?
Is it not strange that two low-precision doubles like this can't divide to the correct value of 28?
No, not really. Neither 0.7 nor 0.025 can be exactly represented in the double type. The exact values involved are:
0.6999999999999999555910790149937383830547332763671875
0.025000000000000001387778780781445675529539585113525390625
Now are you surprised that the division doesn't give exactly 28? Garbage in, garbage out...
As you say, the right result to represent decimal numbers exactly is to use decimal. If the rest of your program is using the wrong type, that just means you need to work out which is higher: the cost of getting the wrong answer, or the cost of changing the whole program.
Precision is always a problem, in case you are dealing with float or double.
Its a known issue in Computer Science and every programming language is affected by it. To minimize these sort of errors, which are mostly related to rounding, a complete field of Numerical Analysis is dedicated to it.
For instance, let take the following code.
What would you expect?
You will expect the answer to be 1, but this is not the case, you will get 0.9999907.
float v = .001f;
float sum = 0;
for (int i = 0; i < 1000; i++ )
{
sum += v;
}
It has nothing to do with how 'simple' or 'small' the double numbers are. Strictly speaking, neither 0.7 or 0.025 may be stored as exactly those numbers in computer memory, so performing calculations on them may provide interesting results if you're after heavy precision.
So yes, use decimal or round.
To explain this by analogy:
Imagine that you are working in base 3. In base 3, 0.1 is (in decimal) 1/3, or 0.333333333'.
So you can EXACTLY represent 1/3 (decimal) in base 3, but you get rounding errors when trying to express it in decimal.
Well, you can get exactly the same thing with some decimal numbers: They can be exactly expressed in decimal, but they CAN'T be exactly expressed in binary; hence, you get rounding errors with them.
Short answer to your first question: No, it's not strange. Floating-point numbers are discrete approximations of the real numbers, which means that rounding errors will propagate and scale when you do arithmetic operations.
Theres' a whole field of mathematics called numerical analyis that basically deal with how to minimize the errors when working with such approximations.
It's the usual floating point imprecision. Not every number can be represented as a double, and those minor representation inaccuracies add up. It's also a reason why you should not compare doubles to exact numbers. I just tested it, and result.ToString() showed 28 (maybe some kind of rounding happens in double.ToString()?). result == 28 returned false though. And (int)result returned 27. So you'll just need to expect imprecisions like that.
A number can have multiple representations if we use a float, so the results of a division of floats may produce bitwise different floats. But what if the denominator is a power of 2?
AFAIK, dividing by a power of 2 would only shift the exponent, leaving the same mantissa, always producing bitwise identical floats. Is that right?
float a = xxx;
float result = n/1024f; // always the same result?
--- UPDATE ----------------------
Sorry for my lack of knowledge in the IEEE black magic for floating points :) , but I'm talking about those numbers Guvante mentioned: no representation for certain decimal numbers, 'inaccurate' floats. For the rest of this post I'll use 'accurate' and 'inaccurate' considering Guvante's definition of these words.
To simplify, let's say the numerator is always an 'accurate' number. Also, let's divide not by any power of 2, but always for 1024. Additionally, I'm doing the operation the same way every time (same method), so I'm talking about getting the same results in different executions (for the same inputs, sure).
I'm asking all this because I see different numbers coming from the same inputs, so I thought: well if I only use 'accurate' floats as numerators and divide by 1024 I will only shift the exponent, still having an 'accurate' float.
You asked for an example. The real problem is this: I have a simulator producing sometimes 0.02999994 and sometimes 0.03000000 for the same inputs. I thought I could multiply these numbers by 1024, round to get an 'integer' ('accurate' float) that would be the same for those two numbers, and then divide by 1024 to get an 'accurate' rounded float.
I was told (in my other question) that I could convert to decimal, round and cast to float, but I want to know if this way works.
A number can have multiple representations if we use a float
The question appears to be predicated on an incorrect premise; the only number that has multiple representations as a float is zero, which can be represented as either "positive zero" or "negative zero". Other than zero a given number only has one representation as a float, assuming that you are talking about the "double" or "float" types.
Or perhaps I misunderstand. Is the issue that you are referring to the fact that the compiler is permitted to do floating point operations in higher precision than the 32 or 64 bits available for storage? That can cause divisions and multiplications to produce different results in some cases.
Since people often don't fully grasp floating point numbers I will go over some of your points real quick. Each particular combination of bits in a floating point number represent a unique number. However because that number has a base 2 fractional component, there is no representation for certain decimal numbers. For instance 1.1. In those cases you take the closest number. IEEE 754-2008 specifies round to nearest, ties to even in these cases.
The real difficulty is when you combine two of these 'inaccurate' numbers. This can introduce problems as each intermediate step will involve rounding. If you calculate the same value using two different methods, you could come up with subtly different values. Typically this is handled with an epsilon when you want equality.
Now onto your real question, can you divide by a power of two and avoid introducing any additional 'inaccuracies'? Normally you can, however as with all floating point numbers, denormals and other odd cases have their own logic, and obviously if your mantissa overflows you will have difficulty. And again note, that no mathematical errors are introduced during any of this, it is simply math being done with limited percision, which involves intermittent rounding of results.
EDIT: In response to new question
What you are saying could work, but is pretty much equivalent to rounding. Additionally if you are just looking for equality, you should use an episilon as I mentioned earlier (a - b) < e for some small value e (0.0001 would work in your example). If you are looking to print out a pretty number, and the framework you are using isn't doing it to your liking, some rounding would be the most direct way of describing your solution, which is always a plus.
Why can't c# do any exact operations.
Math.Pow(Math.Sqrt(2.0),2) == 2.0000000000000004
I know how doubles work, I know where the rounding error is from, I know that it's almost the correct value, and I know that you can't store infinite numbers in a finite double. But why isn't there a way that c# can calculate it exactly, while my calculator can do it.
Edit
It's not about my calculator, I was just giving an example:
http://www.wolframalpha.com/input/?i=Sqrt%282.000000000000000000000000000000000000000000000000000000000000000000000000000000001%29%5E2
Cheers
Chances are your calculator can't do it exactly - but it's probably storing more information than it's displaying, so the error after squaring ends up outside the bounds of what's displayed. Either that, or its errors happen to cancel out in this case - but that's not the same as getting it exactly right in a deliberate way.
Another option is that the calculator is remembering the operations that resulted in the previous results, and applying algebra to cancel out the operations... that seems pretty unlikely though. .NET certainly won't try to do that - it will calculate the intermediate value (the root of two) and then square it.
If you think you can do any better, I suggest you try writing out the square root of two to (say) 50 decimal places, and then square it exactly. See whether you come out with exactly 2...
Your calculator is not calculating it exactly, it just that the rounding error is so small that it's not displayed.
I believe most calculators use binary-coded decimals, which is the equivalent of C#'s decimal type (and thus is entirely accurate). That is, each byte contains two digits of the number and maths is done via logarithms.
What makes you think your calculator can do it? It's almost certainly displaying less digits than it calculates with and you'd get the 'correct' result if you printed out your 2.0000000000000004 with only five fractional digits (for example).
I think you'll probably find that it can't. When I do the square root of 2 and then multiply that by itself, I get 1.999999998.
The square root of 2 is one of those annoying irrational numbers like PI and therefore can't be represented with normal IEEE754 doubles or even decimal types. To represent it exactly, you need a system capable of symbolic math where the value is stored as "the square root of two" so that subsequent calculations can deliver correct results.
The way calculators round up numbers vary from model to model. My TI Voyage 200 does algebra to simplify equations (among other things) but most calculators will display only a portion of the real value calculated, after applying a round function on the result. For example, you may find the square root of 2 and the calculator would store (let's say) 54 decimals, but will only display 12 rounded decimals. Thus when doing a square root of 2, then do a power of that result by 2 would return the same value since the result is rounded. In any case, unless the calculator can keep an infinite number of decimals, you'll always have a best approximate result from complexe operations.
By the way, try to represent 10.0 in binary and you'll realize that you can't represent it evenly and you'll end up with (something like) 10.00000000000..01
Your calculator has methods which recognize and manipulate irrational input values.
For example: 2^(1/2) is likely not evaluated to a number in the calculator if you do not explicitly tell it to do so (as in the ti89/92).
Additionally, the calculator has logic it can use to manipulate them such as x^(1/2) * y^(1/2) = (x*y)^1/2 where it can then wash, rinse, repeat the method for working with irrational values.
If you were to give c# some method to do this, I suppose it could as well. After all, algebraic solvers such as mathematica are not magical.
It has been mentioned before, but I think what you are looking for is a computer algebra system. Examples of these are Maxima and Mathematica, and they are designed solely to provide exact values to mathematical calculations, something not covered by the CPU.
The mathematical routines in languages like C# are designed for numerical calculations: it is expected that if you are doing calculations as a program you will have simplified it already, or you will only need a numerical result.
2.0000000000000004 and 2. are both represented as 10. in single precision. In your case, using single precision for C# should give the exact answer
For your other example, Wolfram Alpha may use higher precision than machine precision for calculation. This adds a big performance penalty. For instance, in Mathematica, going to higher precision makes calculations about 300 times slower
k = 1000000;
vec1 = RandomReal[1, k];
vec2 = SetPrecision[vec1, 20];
AbsoluteTiming[vec1^2;]
AbsoluteTiming[vec2^2;]
It's 0.01 second vs 3 seconds on my machine
You can see the difference in results using single precision and double precision introduced by doing something like the following in Java
public class Bits {
public static void main(String[] args) {
double a1=2.0;
float a2=(float)2.0;
double b1=Math.pow(Math.sqrt(a1),2);
float b2=(float)Math.pow(Math.sqrt(a2),2);
System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(a1)));
System.out.println(Integer.toBinaryString(Float.floatToRawIntBits(a2)));
System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(b1)));
System.out.println(Integer.toBinaryString(Float.floatToRawIntBits(b2)));
}
}
You can see that single precision result is exact, whereas double precision is off by one bit
I always tell in c# a variable of type double is not suitable for money. All weird things could happen. But I can't seem to create an example to demonstrate some of these issues. Can anyone provide such an example?
(edit; this post was originally tagged C#; some replies refer to specific details of decimal, which therefore means System.Decimal).
(edit 2: I was specific asking for some c# code, so I don't think this is language agnostic only)
Very, very unsuitable. Use decimal.
double x = 3.65, y = 0.05, z = 3.7;
Console.WriteLine((x + y) == z); // false
(example from Jon's page here - recommended reading ;-p)
You will get odd errors effectively caused by rounding. In addition, comparisons with exact values are extremely tricky - you usually need to apply some sort of epsilon to check for the actual value being "near" a particular one.
Here's a concrete example:
using System;
class Test
{
static void Main()
{
double x = 0.1;
double y = x + x + x;
Console.WriteLine(y == 0.3); // Prints False
}
}
Yes it's unsuitable.
If I remember correctly double has about 17 significant numbers, so normally rounding errors will take place far behind the decimal point. Most financial software uses 4 decimals behind the decimal point, that leaves 13 decimals to work with so the maximum number you can work with for single operations is still very much higher than the USA national debt. But rounding errors will add up over time. If your software runs for a long time you'll eventually start losing cents. Certain operations will make this worse. For example adding large amounts to small amounts will cause a significant loss of precision.
You need fixed point datatypes for money operations, most people don't mind if you lose a cent here and there but accountants aren't like most people..
edit
According to this site http://msdn.microsoft.com/en-us/library/678hzkk9.aspx Doubles actually have 15 to 16 significant digits instead of 17.
#Jon Skeet decimal is more suitable than double because of its higher precision, 28 or 29 significant decimals. That means less chance of accumulated rounding errors becoming significant. Fixed point datatypes (ie integers that represent cents or 100th of a cent like I've seen used) like Boojum mentions are actually better suited.
Since decimal uses a scaling factor of multiples of 10, numbers like 0.1 can be represented exactly. In essence, the decimal type represents this as 1 / 10 ^ 1, whereas a double would represent this as 104857 / 2 ^ 20 (in reality it would be more like really-big-number / 2 ^ 1023).
A decimal can exactly represent any base 10 value with up to 28/29 significant digits (like 0.1). A double can't.
My understanding is that most financial systems express currency using integers -- i.e., counting everything in cents.
IEEE double precision actually can represent all integers exactly in the range -2^53 through +2^53. (Hacker's Delight, pg. 262) If you use only addition, subtraction and multiplication, and keep everything to integers within this range then you should see no loss of precision. I'd be very wary of division or more complex operations, however.
Using double when you don't know what you are doing is unsuitable.
"double" can represent an amount of a trillion dollars with an error of 1/90th of a cent. So you will get highly precise results. Want to calculate how much it costs to put a man on Mars and get him back alive? double will do just fine.
But with money there are often very specific rules saying that a certain calculation must give a certain result and no other. If you calculate an amount that is very very very close to $98.135 then there will often be a rule that determines whether the result should be $98.14 or $98.13 and you must follow that rule and get the result that is required.
Depending on where you live, using 64 bit integers to represent cents or pennies or kopeks or whatever is the smallest unit in your country will usually work just fine. For example, 64 bit signed integers representing cents can represent values up to 92,223 trillion dollars. 32 bit integers are usually unsuitable.
No a double will always have rounding errors, use "decimal" if you're on .Net...
Actually floating-point double is perfectly well suited to representing amounts of money as long as you pick a suitable unit.
See http://www.idinews.com/moneyRep.html
So is fixed-point long. Either consumes 8 bytes, surely preferable to the 16 consumed by a decimal item.
Whether or not something works (i.e. yields the expected and correct result) is not a matter of either voting or individual preference. A technique either works or it doesn't.