double r = 11.631;
double theta = 21.4;
In the debugger, these are shown as 11.631000000000000 and 21.399999618530273.
How can I avoid this?
These accuracy problems are due to the internal representation of floating point numbers and there's not much you can do to avoid it.
By the way, printing these values at run-time often still leads to the correct results, at least using modern C++ compilers. For most operations, this isn't much of an issue.
I liked Joel's explanation, which deals with a similar binary floating point precision issue in Excel 2007:
See how there's a lot of 0110 0110 0110 there at the end? That's because 0.1 has no exact representation in binary... it's a repeating binary number. It's sort of like how 1/3 has no representation in decimal. 1/3 is 0.33333333 and you have to keep writing 3's forever. If you lose patience, you get something inexact.
So you can imagine how, in decimal, if you tried to do 3*1/3, and you didn't have time to write 3's forever, the result you would get would be 0.99999999, not 1, and people would get angry with you for being wrong.
If you have a value like:
double theta = 21.4;
And you want to do:
if (theta == 21.4)
{
}
You have to be a bit clever, you will need to check if the value of theta is really close to 21.4, but not necessarily that value.
if (fabs(theta - 21.4) <= 1e-6)
{
}
This is partly platform-specific - and we don't know what platform you're using.
It's also partly a case of knowing what you actually want to see. The debugger is showing you - to some extent, anyway - the precise value stored in your variable. In my article on binary floating point numbers in .NET, there's a C# class which lets you see the absolutely exact number stored in a double. The online version isn't working at the moment - I'll try to put one up on another site.
Given that the debugger sees the "actual" value, it's got to make a judgement call about what to display - it could show you the value rounded to a few decimal places, or a more precise value. Some debuggers do a better job than others at reading developers' minds, but it's a fundamental problem with binary floating point numbers.
Use the fixed-point decimal type if you want stability at the limits of precision. There are overheads, and you must explicitly cast if you wish to convert to floating point. If you do convert to floating point you will reintroduce the instabilities that seem to bother you.
Alternately you can get over it and learn to work with the limited precision of floating point arithmetic. For example you can use rounding to make values converge, or you can use epsilon comparisons to describe a tolerance. "Epsilon" is a constant you set up that defines a tolerance. For example, you may choose to regard two values as being equal if they are within 0.0001 of each other.
It occurs to me that you could use operator overloading to make epsilon comparisons transparent. That would be very cool.
For mantissa-exponent representations EPSILON must be computed to remain within the representable precision. For a number N, Epsilon = N / 10E+14
System.Double.Epsilon is the smallest representable positive value for the Double type. It is too small for our purpose. Read Microsoft's advice on equality testing
I've come across this before (on my blog) - I think the surprise tends to be that the 'irrational' numbers are different.
By 'irrational' here I'm just referring to the fact that they can't be accurately represented in this format. Real irrational numbers (like π - pi) can't be accurately represented at all.
Most people are familiar with 1/3 not working in decimal: 0.3333333333333...
The odd thing is that 1.1 doesn't work in floats. People expect decimal values to work in floating point numbers because of how they think of them:
1.1 is 11 x 10^-1
When actually they're in base-2
1.1 is 154811237190861 x 2^-47
You can't avoid it, you just have to get used to the fact that some floats are 'irrational', in the same way that 1/3 is.
One way you can avoid this is to use a library that uses an alternative method of representing decimal numbers, such as BCD
If you are using Java and you need accuracy, use the BigDecimal class for floating point calculations. It is slower but safer.
Seems to me that 21.399999618530273 is the single precision (float) representation of 21.4. Looks like the debugger is casting down from double to float somewhere.
You cant avoid this as you're using floating point numbers with fixed quantity of bytes. There's simply no isomorphism possible between real numbers and its limited notation.
But most of the time you can simply ignore it. 21.4==21.4 would still be true because it is still the same numbers with the same error. But 21.4f==21.4 may not be true because the error for float and double are different.
If you need fixed precision, perhaps you should try fixed point numbers. Or even integers. I for example often use int(1000*x) for passing to debug pager.
Dangers of computer arithmetic
If it bothers you, you can customize the way some values are displayed during debug. Use it with care :-)
Enhancing Debugging with the Debugger Display Attributes
Refer to General Decimal Arithmetic
Also take note when comparing floats, see this answer for more information.
According to the javadoc
"If at least one of the operands to a numerical operator is of type double, then the
operation is carried out using 64-bit floating-point arithmetic, and the result of the
numerical operator is a value of type double. If the other operand is not a double, it is
first widened (§5.1.5) to type double by numeric promotion (§5.6)."
Here is the Source
Related
I am working on a LibGDX program in Java 7 64 bits.
When I use junit to test a function which receives a float as parameter, I got a strange result. I call the function using 123.123456f as parameter, the function receives 123.12346. Why does this happen?
When I use 12.123456f as parameter, it got the correct result.
123.12345f still works.
Hence I use System.out.println(...) to check the input.
It is not important to me, but I just want to know why. Thank you very much!
Regards,
Antony
f means float which means single-precision IEEE-754 floating-point number. They aren't very precise, they only have roughly seven significant digits. You can double that (!) by using d for double, which is a double-precision floating-point number, giving roughly 15 digits of precision. Provided, of course, that whatever you're passing this into accepts doubles and not just floats. If it needs any reasonable precision, it should.
Note, though, that even doubles have issues; they have greater precision, not perfect precision. IEEE-754 floating point is designed for rapid calculation and compact storage. The classic imprecision example, even with doubles, is 0.1 + 0.2, which comes out as 0.30000000000000004.
If you were dealing with currency figures (I think, with that library, you aren't), you might look at BigDecimal, which works more like we're used to working, with an arbitrary number of digits. They're much bigger and much slower, and have their own issues (like the fact they can't accurately represent 1 / 3), but for currency they can be a better choice.
As I tried to debug, I found that : just as I type in
Dim value As Double
value = 0.90000
then hit enter, and it automatically converts to 0.9
Shouldn't it keep the precision in double in visual basic?
For my calculation, I absolutely need to show the precision
If precision is required then the Currency data type is what you want to use.
There are at least two representations of your value in play. One is the value you see on the screen -- a string -- and one is the internal representation -- a binary value. In dealing with fractional values, the two are often not equivalent and where they aren't, it's because they can't be.
If you stick with doubles, VB will maintain 53 bits of mantissa throughout your calculations, no matter how they might appear when printed. If you transition through the string domain, say by saving to a file or DB and later retrieving, it often has to leave some of that precision behind. It's inevitable, because the interface between the two domains is not perfect. Some values that can be exactly represented as strings (or Decimals, that is, powers of ten) can't be exactly represented as fractional powers of 2.
This has nothing to do with VB, it's the nature of floating point. The best you can do is control where the rounding occurs. For this purpose your friend is the Format function, which controls how a value appears in string form.
? Format$(0.9, "0.00000") will show you an example.
You are getting what you see on the screen confused with what bits are being set in the Double to make that number.
VB is simply being "helpful", and simply knocking off excess zeros. But for all intents and purposes,
0.9
is identical to
0.90000
If you don't believe me, try doing this comparison:
Debug.Print CDbl("0.9") = CDbl("0.90000")
As has already been said, displayed precision can be shown using the Format$() function, e.g.
Debug.Print Format$(0.9, "0.00000")
No, it shouldn't keep the precision. Binary floating point values don't retain this information... and it would be somewhat odd to do so, given that you're expressing the value in one base even though it's being represented in another.
I don't know whether VB6 has a decimal floating point type, but that's probably what you want - or a fixed point decimal type, perhaps. Certainly in .NET, System.Decimal has retained extra 0s from .NET 1.1 onwards. If this doesn't help you, you could think about remembering two integers - e.g. "90000" and "100000" in this case, so that the value you're representing is one integer divided by another, with the associated level of precision.
EDIT: I thought that Currency may be what you want, but according to this article, that's fixed at 4 decimal places, and you're trying to retain 5. You could potentially just multiply by 10, if you always want 5 decimal places - but it's an awkward thing to remember to do everywhere... and you'd have to work out how to format it appropriately. It would also always be 4 decimal places, I suspect, even if you'd specified fewer - so if you want "0.300" to be different to "0.3000" then Currency may not be appropriate. I'm entirely basing this on articles online though...
You can also enter the value as 0.9# instead. This helps avoid implicit coercion within an expression that may truncate the precision you expect. In most cases the compiler won't require this hint though because floating point literals default to Double (indeed, the IDE typically deletes the # symbol unless the value was an integer, e.g. 9#).
Contrast the results of these:
MsgBox TypeName(0.9)
MsgBox TypeName(0.9!)
MsgBox TypeName(0.9#)
Why this code 7.30 - 7.20 in ruby returns 0.0999999999999996, not 0.10?
But if i'll write 7.30 - 7.16, for example, everything will be ok, i'll get 0.14.
What the problem, and how can i solve it?
What Every Computer Scientist Should Know About Floating-Point Arithmetic
The problem is that some numbers we can easily write in decimal don't have an exact representation in the particular floating point format implemented by current hardware. A casual way of stating this is that all the integers do, but not all of the fractions, because we normally store the fraction with a 2**e exponent. So, you have 3 choices:
Round off appropriately. The unrounded result is always really really close, so a rounded result is invariably "perfect". This is what Javascript does and lots of people don't even realize that JS does everything in floating point.
Use fixed point arithmetic. Ruby actually makes this really easy; it's one of the only languages that seamlessly shifts to Class Bignum from Fixnum as numbers get bigger.
Use a class that is designed to solve this problem, like BigDecimal
To look at the problem in more detail, we can try to represent your "7.3" in binary. The 7 part is easy, 111, but how do we do .3? 111.1 is 7.5, too big, 111.01 is 7.25, getting closer. Turns out, 111.010011 is the "next closest smaller number", 7.296875, and when we try to fill in the missing .003125 eventually we find out that it's just 111.010011001100110011... forever, not representable in our chosen encoding in a finite bit string.
The problem is that floating point is inaccurate. You can solve it by using Rational, BigDecimal or just plain integers (for example if you want to store money you can store the number of cents as an int instead of the number of dollars as a float).
BigDecimal can accurately store any number that has a finite number of digits in base 10 and rounds numbers that don't (so three thirds aren't one whole).
Rational can accurately store any rational number and can't store irrational numbers at all.
That is a common error from how float point numbers are represented in memory.
Use BigDecimal if you need exact results.
result=BigDecimal.new("7.3")-BigDecimal("7.2")
puts "%2.2f" % result
It is interesting to note that a number that has few decimals in one base may typically have a very large number of decimals in another. For instance, it takes an infinite number of decimals to express 1/3 (=0.3333...) in the base 10, but only one decimal in the base 3. Similarly, it takes many decimals to express the number 1/10 (=0.1) in the base 2.
Since you are doing floating point math then the number returned is what your computer uses for precision.
If you want a closer answer, to a set precision, just multiple the float by that (such as by 100), convert it to an int, do the math, then divide.
There are other solutions, but I find this to be the simplest since rounding always seems a bit iffy to me.
This has been asked before here, you may want to look for some of the answers given before, such as this one:
Dealing with accuracy problems in floating-point numbers
I am writing code that will deal with currencies, charges, etc. I am going to use the BigDecimal class for math and storage, but we ran into something weird with it.
This statement:
1876.8 == BigDecimal('1876.8')
returns false.
If I run those values through a formatting string "%.13f" I get:
"%.20f" % 1876.8 => 1876.8000000000000
"%.20f" % BigDecimal('1876.8') => 1876.8000000000002
Note the extra 2 from the BigDecimal at the last decimal place.
I thought BigDecimal was supposed to counter the inaccuracies of storing real numbers directly in the native floating point of the computer. Where is this 2 coming from?
It won't give you as much control over the number of decimal places, but the conventional format mechanism for BigDecimal appears to be:
a.to_s('F')
If you need more control, consider using the Money gem, assuming your domain problem is mostly about currency.
gem install money
You are right, BigDecimal should be storing it correctly, my best guess is:
BigDecimal is storing the value correctly
When passed to a string formatting function, BigDecimal is being cast as a lower precision floating point value, creating the ...02.
When compared directly with a float, the float has an extra decimal place far beyond the 20 you see (classic floats can't be compared behavoir).
Either way, you are unlikely to get accurate results comparing a float to a BigDecimal.
Don't compare FPU decimal string fractions for equality
The problem is that the equality comparison of a floating or double value with a decimal constant that contains a fraction is rarely successful.
Very few decimal string fractions have exact values in the binary FP representation, so equality comparisons are usually doomed.*
To answer your exact question, the 2 is coming from a slightly different conversion of the decimal string fraction into the Float format. Because the fraction cannot be represented exactly, it's possible that two computations will consider different amounts of precision in intermediate calculations and ultimately end up rounding the result to a 52-bit IEEE 754 double precision mantissa differently. It hardly matters because there is no exact representation anyway, but one is probably more wrong than the other.
In particular, your 1876.8 cannot be represented exactly by an FP object, in fact, between 0.01 and 0.99, only 0.25, 0.50, and 0.75 have exact binary representations. All the others, include 1876.8, repeat forever and are rounded to 52 bits. This is about half of the reason that BigDecimal even exists. (The other half of the reason is the fixed precision of FP data: sometimes you need more.)
So, the result that you get when comparing an actual machine value with a decimal string constant depends on every single bit in the binary fraction ... down to 1/252 ... and even then requires rounding.
If there is anything even the slightest bit (hehe, bit, sorry) imperfect about the process that produced the number, or the input conversion code, or anything else involved, they won't look exactly equal.
An argument could even be made that the comparison should always fail because no IEEE-format FPU can even represent that number exactly. They really are not equal, even though they look like it. On the left, your decimal string has been converted to a binary string, and most of the numbers just don't convert exactly. On the right, it's still a decimal string.
So don't mix floats with BigDecimal, just compare one BigDecimal with another BigDecimal. (Even when both operands are floats, testing for equality requires great care or a fuzzy test. Also, don't trust every formatted digit: output formatting will carry remainders way off the right side of the fraction, so you don't generally start seeing zeroes, you will just see garbage values.)
*The problem: machine numbers are x/2n, but decimal constants are x/(2n * 5m). Your value as sign, exponent, and mantissa is the infinitely repeating 0 10000001001 1101010100110011001100110011001100110011001100110011... Ironically, FP arithmetic is perfectly precise and equality comparisons work perfectly well when the value has no fraction.
as David said, BigDecimal is storing it right
p (BigDecimal('1876.8') * 100000000000000).to_i
returns 187680000000000000
so, yes, the string formatting is ruining it
If you don't need fractional cents, consider storing and manipulating the currency as an integer, then dividing by 100 when it's time to display. I find that easier than dealing with the inevitable precision issues of storing and manipulating in floating point.
On Mac OS X, I'm running ruby 1.8.7 (2008-08-11 patchlevel 72) [i686-darwin9]
irb(main):004:0> 1876.8 == BigDecimal('1876.8') => true
However, being Ruby, I think you should think in terms of messages sent to objects. What does this return to you:
BigDecimal('1876.8') == 1876.8
The two aren't equivalent, and if you're trying to use BigDecimal's ability to determine precise decimal equality, it should be the receiver of the message asking about the equality.
For the same reason I don't think formatting the BigDecimal by sending a format message to the format string is the right approach either.
As I started coding my first app I used NSNumber for money values without thinking twice. Then I thought that maybe c types were enough to deal with my values. Yet, I was advised in the iPhone SDK forum to use NSDecimalNumber, because of its excellent rounding capabilities.
Not being a mathematician by temperament, I thought that the mantissa/exponent paradigm might be overkill; still, googlin' around, I realised that most talks about money/currency in cocoa were referred to NSDecimalNumber.
Notice that the app I am working on is going to be internationalised, so the option of counting the amount in cents is not really viable, for the monetary structure depends greatly on the locale used.
I am 90% sure that I need to go with NSDecimalNumber, but since I found no unambiguous answer on the web (something like: "if you deal with money, use NSDecimalNumber!") I thought I'd ask here. Maybe the answer is obvious to most, but I want to be sure before starting a massive re-factoring of my app.
Convince me :)
Marcus Zarra has a pretty clear stance on this: "If you are dealing with currency at all, then you should be using NSDecimalNumber." His article inspired me to look into NSDecimalNumber, and I've been very impressed with it. IEEE floating point errors when dealing with base-10 math have been irritating me for a while (1 * (0.5 - 0.4 - 0.1) = -0.00000000000000002776) and NSDecimalNumber does away with them.
NSDecimalNumber doesn't just add another few digits of binary floating point precision, it actually does base-10 math. This gets rid of the errors like the one shown in the example above.
Now, I'm writing a symbolic math application, so my desire for 30+ decimal digit precision and no weird floating point errors might be an exception, but I think it's worth looking at. The operations are a little more awkward than simple var = 1 + 2 style math, but they're still manageable. If you're worried about allocating all sorts of instances during your math operations, NSDecimal is the C struct equivalent of NSDecimalNumber and there are C functions for doing the exact same math operations with it. In my experience, these are plenty fast for all but the most demanding applications (3,344,593 additions/s, 254,017 divisions/s on a MacBook Air, 281,555 additions/s, 12,027 divisions/s on an iPhone).
As an added bonus, NSDecimalNumber's descriptionWithLocale: method provides a string with a localized version of the number, including the correct decimal separator. The same goes in reverse for its initWithString:locale: method.
Yes. You have to use
NSDecimalNumber and
not double or float when you deal with currency on iOS.
Why is that??
Because we don't want to get things like $9.9999999998 instead of $10
How that happens??
Floats and doubles are approximations. They always comes with a rounding error. The format computers use to store decimals cause this rouding error.
If you need more details read
http://floating-point-gui.de/
According to apple docs,
NSDecimalNumber is an immutable subclass of NSNumber, provides an object-oriented wrapper for doing base-10 arithmetic. An instance can represent any number that can be expressed as mantissa x 10^exponent where mantissa is a decimal integer up to 38 digits long, and exponent is an integer from –128 through 127.wrapper for doing base-10 arithmetic.
So NSDecimalNumber is recommonded for deal with currency.
(Adapted from my comment on the other answer.)
Yes, you should. An integral number of pennies works only as long as you don't need to represent, say, half a cent. If that happens, you could change it to count half-cents, but what if you then need to represent a quarter-cent, or an eighth of a cent?
The only proper solution is NSDecimalNumber (or something like it), which puts off the problem to 10^-128¢ (i.e.,
0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000001¢).
(Another way would be arbitrary-precision arithmetic, but that requires a separate library, such as the GNU MP Bignum library. GMP is under the LGPL. I've never used that library and don't know exactly how it works, so I couldn't say how well it would work for you.)
[Edit: Apparently, at least one person—Brad Larson—thinks I'm talking about binary floating-point somewhere in this answer. I'm not.]
I've found it convenient to use an integer to represent the number of cents and then divide by 100 for presentation. Avoids the whole issue.
A better question is, when should you not use NSDecimalNumber to deal with money. The short answer to that question is, when you can't tolerate the performance overhead of NSDecimalNumber and you don't care about small rounding errors because you're never dealing with more than a few digits of precision. The even shorter answer is, you should always use NSDecimalNumber when dealing with money.
VISA, MasterCards and others are using integer values while passing amounts. It's up to sender and reciever to parse amouts correctly according to currency exponent (divide or multiply by 10^num, where num - is an exponent of the currency). Note that different currencies have different exponents. Usually it's 2 (hence we divide and multiply by 100), but some currencies have exponent = 0 (VND,etc), or = 3.