What would you use if you had to hold numbers bigger than the ulong type can hold (0 to 18,446,744,073,709,551,615)? For instance measuring distance between planets or galaxies?
For values which only require up to 28 digits, you can use System.Decimal. The way it's designed, you don't encounter the scaling issue you do with double, where for large numbers the gap between two adjacent numbers is bigger than 1. Admittedly this looks slightly odd, given that decimal is usually used for non-integer values - but in some cases I think it's a perfectly reasonable use, so long as you document it properly.
For values bigger than that, you can use System.Numerics.BigInteger.
Another alternative is to just use double and accept that you really don't get a precision down to the integer. When it comes to the distance between galaxies, are you really going to have a value which is accurate to a metre anyway? It does depend on how you're going to use this - it can certainly make testing simpler if you use a nicely-predictable integer type, but you should really think about where the values are going to go and what you're going to do with them.
I rather doubt you need to have integer types to store something like the distance between the galaxies. To my knowledge very large integer values are required in few cases like (demonstrative) encryption. For these cases, you can use System.Numerics.BigInteger.
I think you should use floating point types (e.g. double) for purposes like one you described.
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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 want my cake and to eat it. I want to beautify (round) numbers to the largest extent possible without compromising accuracy for other calculations. I'm using doubles in C# (with some string conversion manipulation too).
Here's the issue. I understand the inherent limitations in double number representation (so please don't explain that). HOWEVER, I want to round the number in some way to appear aesthetically pleasing to the end user (I am making a calculator). The problem is rounding by X significant digits works in one case, but not in the other, whilst rounding by decimal place works in the other, but not the first case.
Observe:
CASE A: Math.Sin(Math.Pi) = 0.000000000000000122460635382238
CASE B: 0.000000000000001/3 = 0.000000000000000333333333333333
For the first case, I want to round by DECIMAL PLACES. That would give me the nice neat zero I'm looking for. Rounding by Sig digits would mean I would keep the erroneous digits too.
However for the second case, I want to round by SIGNIFICANT DIGITS, as I would lose tons of accuracy if I rounded merely by decimal places.
Is there a general way I can cater to both types of calculation?
I don't thinks it's feasible to do that to the result itself and precision has nothing to do with it.
Consider this input: (1+3)/2^3 . You can "beautify" it by showing the result as sin(30) or cos(60) or 1/2 and a whole lot of other interpretations. Choosing the wrong "beautification" can mislead your user, making them think their function has something to do with sin(x).
If your calculator keeps all the initial input as variables you could keep all the operations postponed until you need the result and then make sure you simplify the result until it matches your needs. And you'll need to consider using rational numbers, e, Pi and other irrational numbers may not be as easy to deal with.
The best solution to this is to keep every bit you can get during calculations, and leave the display format up to the end user. The user should have some idea how many significant digits make sense in their situation, given both the nature of the calculations and the use of the result.
Default to a reasonable number of significant digits for a few calculations in the floating point format you are using internally - about 12 if you are using double. If the user changes the format, immediately redisplay in the new format.
The best solution is to use arbitrary-precision and/or symbolic arithmetic, although these result in much more complex code and slower speed. But since performance isn't important for a calculator (in case of a button calculator and not the one that you enter expressions to calculate) you can use them without issue
Anyway there's a good trade-off which is to use decimal floating point. You'll need to limit the input/output precision but use a higher precision for the internal representation so that you can discard values very close to zero like the sin case above. For better results you could detect some edge cases such as sine/cosine of 45 degree's multiples... and directly return the exact result.
Edit: just found a good solution but haven't had an opportunity to try.
Here’s something I bet you never think about, and for good reason: how are floating-point numbers rendered as text strings? This is a surprisingly tough problem, but it’s been regarded as essentially solved since about 1990.
Prior to Steele and White’s "How to print floating-point numbers accurately", implementations of printf and similar rendering functions did their best to render floating point numbers, but there was wide variation in how well they behaved. A number such as 1.3 might be rendered as 1.29999999, for instance, or if a number was put through a feedback loop of being written out and its written representation read back, each successive result could drift further and further away from the original.
...
In 2010, Florian Loitsch published a wonderful paper in PLDI, "Printing floating-point numbers quickly and accurately with integers", which represents the biggest step in this field in 20 years: he mostly figured out how to use machine integers to perform accurate rendering! Why do I say "mostly"? Because although Loitsch's "Grisu3" algorithm is very fast, it gives up on about 0.5% of numbers, in which case you have to fall back to Dragon4 or a derivative
Here be dragons: advances in problems you didn’t even know you had
All experienced programmers in C# (I think this comes from C) are used to cast on of the integers in a division to get the decimal / double / float result instead of the int (the real result truncated).
I'd like to know why is this implemented like this? Is there ANY good reason to truncate the result if both numbers are integer?
C# traces its heritage to C, so the answer to "why is it like this in C#?" is a combination of "why is it like this in C?" and "was there no good reason to change?"
The approach of C is to have a fairly close correspondence between the high-level language and low-level operations. Processors generally implement integer division as returning a quotient and a remainder, both of which are of the same type as the operands.
(So my question would be, "why doesn't integer division in C-like languages return two integers", not "why doesn't it return a floating point value?")
The solution was to provide separate operations for division and remainder, each of which returns an integer. In the context of C, it's not surprising that the result of each of these operations is an integer. This is frequently more accurate than floating-point arithmetic. Consider the example from your comment of 7 / 3. This value cannot be represented by a finite binary number nor by a finite decimal number. In other words, on today's computers, we cannot accurately represent 7 / 3 unless we use integers! The most accurate representation of this fraction is "quotient 2, remainder 1".
So, was there no good reason to change? I can't think of any, and I can think of a few good reasons not to change. None of the other answers has mentioned Visual Basic which (at least through version 6) has two operators for dividing integers: / converts the integers to double, and returns a double, while \ performs normal integer arithmetic.
I learned about the \ operator after struggling to implement a binary search algorithm using floating-point division. It was really painful, and integer division came in like a breath of fresh air. Without it, there was lots of special handling to cover edge cases and off-by-one errors in the first draft of the procedure.
From that experience, I draw the conclusion that having different operators for dividing integers is confusing.
Another alternative would be to have only one integer operation, which always returns a double, and require programmers to truncate it. This means you have to perform two int->double conversions, a truncation and a double->int conversion every time you want integer division. And how many programmers would mistakenly round or floor the result instead of truncating it? It's a more complicated system, and at least as prone to programmer error, and slower.
Finally, in addition to binary search, there are many standard algorithms that employ integer arithmetic. One example is dividing collections of objects into sub-collections of similar size. Another is converting between indices in a 1-d array and coordinates in a 2-d matrix.
As far as I can see, no alternative to "int / int yields int" survives a cost-benefit analysis in terms of language usability, so there's no reason to change the behavior inherited from C.
In conclusion:
Integer division is frequently useful in many standard algorithms.
When the floating-point division of integers is needed, it may be invoked explicitly with a simple, short, and clear cast: (double)a / b rather than a / b
Other alternatives introduce more complication both the programmer and more clock cycles for the processor.
Is there ANY good reason to truncate the result if both numbers are integer?
Of course; I can think of a dozen such scenarios easily. For example: you have a large image, and a thumbnail version of the image which is 10 times smaller in both dimensions. When the user clicks on a point in the large image, you wish to identify the corresponding pixel in the scaled-down image. Clearly to do so, you divide both the x and y coordinates by 10. Why would you want to get a result in decimal? The corresponding coordinates are going to be integer coordinates in the thumbnail bitmap.
Doubles are great for physics calculations and decimals are great for financial calculations, but almost all the work I do with computers that does any math at all does it entirely in integers. I don't want to be constantly having to convert doubles or decimals back to integers just because I did some division. If you are solving physics or financial problems then why are you using integers in the first place? Use nothing but doubles or decimals. Use integers to solve finite mathematics problems.
Calculating on integers is faster (usually) than on floating point values. Besides, all other integer/integer operations (+, -, *) return an integer.
EDIT:
As per the request of the OP, here's some addition:
The OP's problem is that they think of / as division in the mathematical sense, and the / operator in the language performs some other operation (which is not the math. division). By this logic they should question the validity of all other operations (+, -, *) as well, since those have special overflow rules, which is not the same as would be expected from their math counterparts. If this is bothersome for someone, they should find another language where the operations perform as expected by the person.
As for the claim on perfomance difference in favor of integer values: When I wrote the answer I only had "folk" knowledge and "intuition" to back up the claim (hece my "usually" disclaimer). Indeed as Gabe pointed out, there are platforms where this does not hold. On the other hand I found this link (point 12) that shows mixed performances on an Intel platform (the language used is Java, though).
The takeaway should be that with performance many claims and intuition are unsubstantiated until measured and found true.
Yes, if the end result needs to be a whole number. It would depend on the requirements.
If these are indeed your requirements, then you would not want to store a decimal and then truncate it. You would be wasting memory and processing time to accomplish something that is already built-in functionality.
The operator is designed to return the same type as it's input.
Edit (comment response):
Why? I don't design languages, but I would assume most of the time you will be sticking with the data types you started with and in the remaining instance, what criteria would you use to automatically assume which type the user wants? Would you automatically expect a string when you need it? (sincerity intended)
If you add an int to an int, you expect to get an int. If you subtract an int from an int, you expect to get an int. If you multiple an int by an int, you expect to get an int. So why would you not expect an int result if you divide an int by an int? And if you expect an int, then you will have to truncate.
If you don't want that, then you need to cast your ints to something else first.
Edit: I'd also note that if you really want to understand why this is, then you should start looking into how binary math works and how it is implemented in an electronic circuit. It's certainly not necessary to understand it in detail, but having a quick overview of it would really help you understand how the low-level details of the hardware filter through to the details of high-level languages.
Is there a library for decimal calculation, especially the Pow(decimal, decimal) method? I can't find any.
It can be free or commercial, either way, as long as there is one.
Note: I can't do it myself, can't use for loops, can't use Math.Pow, Math.Exp or Math.Log, because they all take doubles, and I can't use doubles. I can't use a serie because it would be as precise as doubles.
One of the multipliyers is a rate : 1/rate^(days/365).
The reason there is no decimal power function is because it would be pointless to use decimal for that calculation. Use double.
Remember, the point of decimal is to ensure that you get exact arithmetic on values that can be exactly represented as short decimal numbers. For reasonable values of rate and days, the values of any of the other subexpressions are clearly not going to be exactly represented as short decimal values. You're going to be dealing with inexact values, so use a type designed for fast calculations of slightly inexact values, like double.
The results when computed in doubles are going to be off by a few billionths of a penny one way or the other. Who cares? You'll round out the error later. Do the rate calculation in doubles. Once you have a result that needs to be turned back into a currency again, multiply the result by ten thousand, round it off to the nearest integer, convert that to a decimal, and then divide it out by ten thousand again, and you'll have a result accurate to four decimal places, which ought to be plenty for a financial calculation.
Here is what I used.
output = (decimal)Math.Pow((double)var1, (double)var2);
Now I'm just learning but this did work but I don't know if I can explain it correctly.
what I believe this does is take the input of var1 and var2 and cast them to doubles to use as the argument for the math.pow method. After that have (decimal) in front of math.pow take the value back to a decimal and place the value in the output variable.
I hope someone can correct me if my explination is wrong but all I know is that it worked for me.
I know this is an old thread but I'm putting this here in case someone finds it when searching for a solution.
If you don't want to mess around with casting and doing you own custom implementation you can install the NuGet DecimalMath.DecimalEx and use it like DecimalEx.Pow(number,power).
Well, here is the Wikipedia page that lists current C# numerics libraries. But TBH I don't think there is a lot of support for decimals
http://en.wikipedia.org/wiki/List_of_numerical_libraries
It's kind of inappropriate to use decimals for this kind of calculation in general. It's high precision yes - but it's also low range. As the MSDN docs state it's for financial/monetary calculations - where there isn't much call for POW unfortunately!
Of course you might have a specific problem domain that needs super high precision and all numbers are within 10(28) - 10(-28). But in that case you will probably just need to write your own series calculator such as the one linked to in the comments to the question.
Not using decimal. Use double instead. According to this thread, the Math.Pow(double, double) is called directly from CLR.
How is Math.Pow() implemented in .NET Framework?
Here is what .NET Framework 4 has (2 lines only)
[SecuritySafeCritical]
public static extern double Pow(double x, double y);
64-bit decimal is not native in this 32-bit CLR yet. Maybe on 64-bit Framework in the future?
wait, huh? why can't you use doubles? you could always cast if you're using ints or something:
int a = 1;
int b = 2;
int result = (int)Math.Pow(a,b);
Dividing an int by zero, will throw an exception, but a float won't - at least in Java. Why does a float have additional NaN info, while an int type doesn't?
The representation of a float has been designed such that there are some special combination of bits reserved to store special values such as NaN, infinity, etc.
There are no unused representations for an int type - every bit pattern corresponds to an integer. This has many advantages:
The range of an integer type is as large as possible - no bit patterns are wasted.
The representation of an integer is easy to understand because there are no special cases.
Integer arithmetic can be done at extremely high speed even on very simple processors.
A clear Explanation about float arithmetic is given here
http://www.artima.com/underthehood/floatingP.html
I think the real reason, the root of this, is the well known fact: computers store everything in zeroes and ones.
What does it have to do with integers, floats and zero division? It's pretty simple. If you have only zeroes and ones, it is pretty easy to combine them into integer numbers, like you do with decimal digits. So "10" becomes two, "11" becomes three and so on. This kind of integer representation is so natural that no one would think of inventing anything else for integers, it would just make CPUs more complicated and things more confusing. The only "invention" that was required is to figure out how to store negative numbers, but that's also very natural and simple if you start from the point that x+(-x) should always be equal to zero, without using any special kind of addition here. That's why 11111111 is -1 for 8-bit integers, because if you add 1 to it, it becomes 100000000, then 8th bit is truncated due to overflow and you get your zero. But this natural format has no place for infinities and NaNs, and nobody wanted to invent a non-natural representation just for that. Well, I won't be surprised if someone actually did that, but there is no way such format would become well-known and widely used.
Now, for floating-point numbers, there is no natural representation. Even if we translate 0.5 to binary, it would still be something like 0.1 only now we have "binary point" instead of decimal point. But CPUs can't naturally represent a "point", only 1 and 0. So some kind of special format was needed. There was simply no other way to go. And then someone probably suggested, "Hey guys, while we are at it, why not to include special representation for infinity and other numeric nonsense?" and so it was done.
This is the reason why these formats are so different. How to handle divisions by zero, it's up to language designers, but for floating-points they have the choice between inf/NaN and exceptions, while for integers they don't naturally have such kind of thing.
Basically, it's a purely arbitrary decision.
The traditional int tries to use all the bits for representing possible numbers, whereas IEEE 754 standard reserves a special value for NaN.
The standard could be changed for ints to include special values, at a cost of less efficient operations. The developers usually expect int operations to be very efficient, whereas the operations with floating point numbers are (purely psychologically) more allowed to be slower.
Ints and floats are represented differently inside the machine. Integers usually use a signed, two's complement representation that is (essentially) the number written out in base two. Floats, on the other hand, use a more complex representation that can hold much larger and much smaller values. However, the machine reserves several special bit patterns for floats to mean things other than numbers. There's values for NaN, and for positive or negative infinity, for example. This means that if you divide a float by zero, there is a series of bits that the computer can use to encode that you divided by zero. For ints, all bit patterns are used to encode numbers, so there's no meaningful series of bits the computer could use to represent the error.
This isn't an essential property of ints, though. One could, in theory, make an integer representation that handles division by zero by returning some NaN variant. It's just not what's done in practice.
Java reflects the way most CPUs are implemented. Integer divide by zero causes an interrupt on x86/x64 and Floating point divide by zero results in Infinity, Negative infinity or NaN. Note: with floating point you can also divide by negative zero. :P