I've searched half the day and found some very interesting things about using fixed point data types and bit shifting in C++ to accomplish division operations while avoiding floating point math. However, I have only been able to understand a small fraction of it and I can't seem to get anything to work.
All I'm wanting to do is to take two integers, ad them up, and divide by two to get the average. I need to be able to do this very quickly though, since I'm interpolating camera pixel data on an Arduino and I also have other operations to do.
So I'm confused about shifting in general. Say the integer I want to divide by two is 27. Half of 27 is 13.5. But no matter what fixed point datatype I try, I can only get 13 as an output. For example:
uint8_t x = 27;
Serial.println( x >> 1 );
returns 13
There's got to be some simple way to do this, right?
Fixed point does give you a way to represent 13.5. The Wikipedia article on the Q number format is informative: https://en.wikipedia.org/wiki/Q_(number_format)
Think of it this way: You keep using integers, but instead of taking them at face value, divide them all implicitly by a power of 2 to obtain their semantic value.
So, if using an unsigned byte as your base type (values between 0 and 255, inclusive), you might implicitly divide by 2**3 (8). Now to represent 27, you need an integer set to 27*8=>216. To divide by two, you shift one to the right; now your integer is 108, which when divided by the implicit denominator of 8 gives 13.5, the value you're expecting.
You have to realize that fixed-point number systems (and floating point too, though it's less immediately evident) still have limits, of course; certain operations will overflow no matter what you do, and some operations cause a loss of precision. This is a normal consequence of working with limited-size types.
Say the integer I want to divide by two is 27. Half of 27 is 13.5. But
no matter what fixed point data type I try, I can only get 13 as an
output.
From wikipedia Fixed-Point Arithmetic:
The scaling factor is usually a power of 10 (for human convenience) or
a power of 2 (for computational efficiency).
You actually mentioned fixed point data type, and I think that is the best approach. But no matter what you tried? Perhaps we have different understandings of fixed-point-arithmetic.
while avoiding floating point math.
Another worth while goal, though reducing in value. Even in embedded systems, I seldom had to deal with a processor that did not have floating point parts. Floating point hardware has gotten reasonably good.
Any way, using fixed point avoids any need for floating point. Even for display purposes.
I think I need to proceed with a few examples.
Fixed point Example 1: Dollars and pennies
The unit of American money is based on the dollar. The Dollar is a fixed point data type.
So, if you have 27 dollars, how do you split it with your sibling?
One way (of several) that you all know is to convert 27 dollars into 2700 pennies. Dividing this value by 2 is trivial. Now you and your sibling can each get 1350 pennies. (i.e. the penny is a fixed point data type, that easily converts to/from dollars, and vice-vesa)
Note that this is completely integer arithmetic. Adding 2 integers, and dividing by 2 (any modern compiler will choose the fastest implementation.. either integer divide or perhaps right-shift-by-2) and on my desktop these 2 actions take less than a microsecond to complete.
You should waste no more time on measuring the relative performance of those two options (divide vs right-shift), you simply enable -O3 when your code tests correct. Your compiler should be able to choose correctly.
The choice of units in any problem is based on a scale factor that covers the range of values (in your problem) AND the understandable and quickly implemented conversion between units. And note that uint64_t can describe a large amount of cash, even in pennies. (challenge to the student.)
In General, about fixed point:
Given
uint8_t x = 27;
and the desire to divide by 2 evenly and quickly... can any scale factor be something that serves your needs? I say yes.
example 2 - 50 cent coins and a dollar
How about we try, for example, a simple scale factor of 2, i.e. the unit is a hu, or half unit. (analogous to the 50-cent-coin)
uint8_t x = 27 * 1/hu; (hu = 1/2)
This means that 54 hu represents 27 units. (ie, it takes 54 50-cent-coins to add up to 27 dollars)
The fixed point solution is to scale your integer values to achieve the arithmetic required. If you scale to even values, all your integers will divide evenly to the hu units.
example 3 - nickles and a dollar
Another possible scale might be 20, both decimal (for readability) and binary for performance. (note that there are 20 nickels in a dollar)
uint16 x = 27 * 1/tu; (tu = 1/20)
Now 540 represents a scaled 27. i.e. 540 nickles
All examples are fully integer, provide exact answers, and there is a trivial mechanism to convert the values for presentation to the user. i.e. which ever fixed point used, convert to analogue of pennies, and thus 1350 pennies.
Display the penny count as a dollar
std::cout << (pennyCount / 100) << "." << (pennyCount % 100) << std::endl;
I think this should look something like (untested)
13.50
Now your challenge is to make it look nice on the output.
The reason you only get 13 is because you are actually cutting off the least significant bits when you bit shift. Since you are cutting them off, there is no remainder to check. If you are interested in what your remainder is, you could do something like:
uint8_t x = 27;
Serial.println((x - (x >> 1) - (x >> 1));
(x - (x >> 1)) should give 14 here.
it would be pretty simple to add .5 to a number once you determine whether the remainder is 1.
The following should work and should be fast:
float y = (x >> 1) + (0.5 * (x & 0x01))
What it does
(x >> 1) Divide by 2 using the bit shift
(0.5 * (x & 0x01)) Add 0.5 if the last bit was 1 (odd number)
There have been several questions posted to SO about floating-point representation. For example, the decimal number 0.1 doesn't have an exact binary representation, so it's dangerous to use the == operator to compare it to another floating-point number. I understand the principles behind floating-point representation.
What I don't understand is why, from a mathematical perspective, are the numbers to the right of the decimal point any more "special" that the ones to the left?
For example, the number 61.0 has an exact binary representation because the integral portion of any number is always exact. But the number 6.10 is not exact. All I did was move the decimal one place and suddenly I've gone from Exactopia to Inexactville. Mathematically, there should be no intrinsic difference between the two numbers -- they're just numbers.
By contrast, if I move the decimal one place in the other direction to produce the number 610, I'm still in Exactopia. I can keep going in that direction (6100, 610000000, 610000000000000) and they're still exact, exact, exact. But as soon as the decimal crosses some threshold, the numbers are no longer exact.
What's going on?
Edit: to clarify, I want to stay away from discussion about industry-standard representations, such as IEEE, and stick with what I believe is the mathematically "pure" way. In base 10, the positional values are:
... 1000 100 10 1 1/10 1/100 ...
In binary, they would be:
... 8 4 2 1 1/2 1/4 1/8 ...
There are also no arbitrary limits placed on these numbers. The positions increase indefinitely to the left and to the right.
Decimal numbers can be represented exactly, if you have enough space - just not by floating binary point numbers. If you use a floating decimal point type (e.g. System.Decimal in .NET) then plenty of values which can't be represented exactly in binary floating point can be exactly represented.
Let's look at it another way - in base 10 which you're likely to be comfortable with, you can't express 1/3 exactly. It's 0.3333333... (recurring). The reason you can't represent 0.1 as a binary floating point number is for exactly the same reason. You can represent 3, and 9, and 27 exactly - but not 1/3, 1/9 or 1/27.
The problem is that 3 is a prime number which isn't a factor of 10. That's not an issue when you want to multiply a number by 3: you can always multiply by an integer without running into problems. But when you divide by a number which is prime and isn't a factor of your base, you can run into trouble (and will do so if you try to divide 1 by that number).
Although 0.1 is usually used as the simplest example of an exact decimal number which can't be represented exactly in binary floating point, arguably 0.2 is a simpler example as it's 1/5 - and 5 is the prime that causes problems between decimal and binary.
Side note to deal with the problem of finite representations:
Some floating decimal point types have a fixed size like System.Decimal others like java.math.BigDecimal are "arbitrarily large" - but they'll hit a limit at some point, whether it's system memory or the theoretical maximum size of an array. This is an entirely separate point to the main one of this answer, however. Even if you had a genuinely arbitrarily large number of bits to play with, you still couldn't represent decimal 0.1 exactly in a floating binary point representation. Compare that with the other way round: given an arbitrary number of decimal digits, you can exactly represent any number which is exactly representable as a floating binary point.
For example, the number 61.0 has an exact binary representation because the integral portion of any number is always exact. But the number 6.10 is not exact. All I did was move the decimal one place and suddenly I've gone from Exactopia to Inexactville. Mathematically, there should be no intrinsic difference between the two numbers -- they're just numbers.
Let's step away for a moment from the particulars of bases 10 and 2. Let's ask - in base b, what numbers have terminating representations, and what numbers don't? A moment's thought tells us that a number x has a terminating b-representation if and only if there exists an integer n such that x b^n is an integer.
So, for example, x = 11/500 has a terminating 10-representation, because we can pick n = 3 and then x b^n = 22, an integer. However x = 1/3 does not, because whatever n we pick we will not be able to get rid of the 3.
This second example prompts us to think about factors, and we can see that for any rational x = p/q (assumed to be in lowest terms), we can answer the question by comparing the prime factorisations of b and q. If q has any prime factors not in the prime factorisation of b, we will never be able to find a suitable n to get rid of these factors.
Thus for base 10, any p/q where q has prime factors other than 2 or 5 will not have a terminating representation.
So now going back to bases 10 and 2, we see that any rational with a terminating 10-representation will be of the form p/q exactly when q has only 2s and 5s in its prime factorisation; and that same number will have a terminating 2-representatiion exactly when q has only 2s in its prime factorisation.
But one of these cases is a subset of the other! Whenever
q has only 2s in its prime factorisation
it obviously is also true that
q has only 2s and 5s in its prime factorisation
or, put another way, whenever p/q has a terminating 2-representation, p/q has a terminating 10-representation. The converse however does not hold - whenever q has a 5 in its prime factorisation, it will have a terminating 10-representation , but not a terminating 2-representation. This is the 0.1 example mentioned by other answers.
So there we have the answer to your question - because the prime factors of 2 are a subset of the prime factors of 10, all 2-terminating numbers are 10-terminating numbers, but not vice versa. It's not about 61 versus 6.1 - it's about 10 versus 2.
As a closing note, if by some quirk people used (say) base 17 but our computers used base 5, your intuition would never have been led astray by this - there would be no (non-zero, non-integer) numbers which terminated in both cases!
The root (mathematical) reason is that when you are dealing with integers, they are countably infinite.
Which means, even though there are an infinite amount of them, we could "count out" all of the items in the sequence, without skipping any. That means if we want to get the item in the 610000000000000th position in the list, we can figure it out via a formula.
However, real numbers are uncountably infinite. You can't say "give me the real number at position 610000000000000" and get back an answer. The reason is because, even between 0 and 1, there are an infinite number of values, when you are considering floating-point values. The same holds true for any two floating point numbers.
More info:
http://en.wikipedia.org/wiki/Countable_set
http://en.wikipedia.org/wiki/Uncountable_set
Update:
My apologies, I appear to have misinterpreted the question. My response is about why we cannot represent every real value, I hadn't realized that floating point was automatically classified as rational.
To repeat what I said in my comment to Mr. Skeet: we can represent 1/3, 1/9, 1/27, or any rational in decimal notation. We do it by adding an extra symbol. For example, a line over the digits that repeat in the decimal expansion of the number. What we need to represent decimal numbers as a sequence of binary numbers are 1) a sequence of binary numbers, 2) a radix point, and 3) some other symbol to indicate the repeating part of the sequence.
Hehner's quote notation is a way of doing this. He uses a quote symbol to represent the repeating part of the sequence. The article: http://www.cs.toronto.edu/~hehner/ratno.pdf and the Wikipedia entry: http://en.wikipedia.org/wiki/Quote_notation.
There's nothing that says we can't add a symbol to our representation system, so we can represent decimal rationals exactly using binary quote notation, and vice versa.
BCD - Binary-coded Decimal - representations are exact. They are not very space-efficient, but that's a trade-off you have to make for accuracy in this case.
This is a good question.
All your question is based on "how do we represent a number?"
ALL the numbers can be represented with decimal representation or with binary (2's complement) representation. All of them !!
BUT some (most of them) require infinite number of elements ("0" or "1" for the binary position, or "0", "1" to "9" for the decimal representation).
Like 1/3 in decimal representation (1/3 = 0.3333333... <- with an infinite number of "3")
Like 0.1 in binary ( 0.1 = 0.00011001100110011.... <- with an infinite number of "0011")
Everything is in that concept. Since your computer can only consider finite set of digits (decimal or binary), only some numbers can be exactly represented in your computer...
And as said Jon, 3 is a prime number which isn't a factor of 10, so 1/3 cannot be represented with a finite number of elements in base 10.
Even with arithmetic with arbitrary precision, the numbering position system in base 2 is not able to fully describe 6.1, although it can represent 61.
For 6.1, we must use another representation (like decimal representation, or IEEE 854 that allows base 2 or base 10 for the representation of floating-point values)
If you make a big enough number with floating point (as it can do exponents), then you'll end up with inexactness in front of the decimal point, too. So I don't think your question is entirely valid because the premise is wrong; it's not the case that shifting by 10 will always create more precision, because at some point the floating point number will have to use exponents to represent the largeness of the number and will lose some precision that way as well.
It's the same reason you cannot represent 1/3 exactly in base 10, you need to say 0.33333(3). In binary it is the same type of problem but just occurs for different set of numbers.
(Note: I'll append 'b' to indicate binary numbers here. All other numbers are given in decimal)
One way to think about things is in terms of something like scientific notation. We're used to seeing numbers expressed in scientific notation like, 6.022141 * 10^23. Floating point numbers are stored internally using a similar format - mantissa and exponent, but using powers of two instead of ten.
Your 61.0 could be rewritten as 1.90625 * 2^5, or 1.11101b * 2^101b with the mantissa and exponents. To multiply that by ten and (move the decimal point), we can do:
(1.90625 * 2^5) * (1.25 * 2^3) = (2.3828125 * 2^8) = (1.19140625 * 2^9)
or in with the mantissa and exponents in binary:
(1.11101b * 2^101b) * (1.01b * 2^11b) = (10.0110001b * 2^1000b) = (1.00110001b * 2^1001b)
Note what we did there to multiply the numbers. We multiplied the mantissas and added the exponents. Then, since the mantissa ended greater than two, we normalized the result by bumping the exponent. It's just like when we adjust the exponent after doing an operation on numbers in decimal scientific notation. In each case, the values that we worked with had a finite representation in binary, and so the values output by the basic multiplication and addition operations also produced values with a finite representation.
Now, consider how we'd divide 61 by 10. We'd start by dividing the mantissas, 1.90625 and 1.25. In decimal, this gives 1.525, a nice short number. But what is this if we convert it to binary? We'll do it the usual way -- subtracting out the largest power of two whenever possible, just like converting integer decimals to binary, but we'll use negative powers of two:
1.525 - 1*2^0 --> 1
0.525 - 1*2^-1 --> 1
0.025 - 0*2^-2 --> 0
0.025 - 0*2^-3 --> 0
0.025 - 0*2^-4 --> 0
0.025 - 0*2^-5 --> 0
0.025 - 1*2^-6 --> 1
0.009375 - 1*2^-7 --> 1
0.0015625 - 0*2^-8 --> 0
0.0015625 - 0*2^-9 --> 0
0.0015625 - 1*2^-10 --> 1
0.0005859375 - 1*2^-11 --> 1
0.00009765625...
Uh oh. Now we're in trouble. It turns out that 1.90625 / 1.25 = 1.525, is a repeating fraction when expressed in binary: 1.11101b / 1.01b = 1.10000110011...b Our machines only have so many bits to hold that mantissa and so they'll just round the fraction and assume zeroes beyond a certain point. The error you see when you divide 61 by 10 is the difference between:
1.100001100110011001100110011001100110011...b * 2^10b
and, say:
1.100001100110011001100110b * 2^10b
It's this rounding of the mantissa that leads to the loss of precision that we associate with floating point values. Even when the mantissa can be expressed exactly (e.g., when just adding two numbers), we can still get numeric loss if the mantissa needs too many digits to fit after normalizing the exponent.
We actually do this sort of thing all the time when we round decimal numbers to a manageable size and just give the first few digits of it. Because we express the result in decimal it feels natural. But if we rounded a decimal and then converted it to a different base, it'd look just as ugly as the decimals we get due to floating point rounding.
I'm surprised no one has stated this yet: use continued fractions. Any rational number can be represented finitely in binary this way.
Some examples:
1/3 (0.3333...)
0; 3
5/9 (0.5555...)
0; 1, 1, 4
10/43 (0.232558139534883720930...)
0; 4, 3, 3
9093/18478 (0.49209871198181621387596060179673...)
0; 2, 31, 7, 8, 5
From here, there are a variety of known ways to store a sequence of integers in memory.
In addition to storing your number with perfect accuracy, continued fractions also have some other benefits, such as best rational approximation. If you decide to terminate the sequence of numbers in a continued fraction early, the remaining digits (when recombined to a fraction) will give you the best possible fraction. This is how approximations to pi are found:
Pi's continued fraction:
3; 7, 15, 1, 292 ...
Terminating the sequence at 1, this gives the fraction:
355/113
which is an excellent rational approximation.
In the equation
2^x = y ;
x = log(y) / log(2)
Hence, I was just wondering if we could have a logarithmic base system for binary like,
2^1, 2^0, 2^(log(1/2) / log(2)), 2^(log(1/4) / log(2)), 2^(log(1/8) / log(2)),2^(log(1/16) / log(2)) ........
That might be able to solve the problem, so if you wanted to write something like 32.41 in binary, that would be
2^5 + 2^(log(0.4) / log(2)) + 2^(log(0.01) / log(2))
Or
2^5 + 2^(log(0.41) / log(2))
The problem is that you do not really know whether the number actually is exactly 61.0 . Consider this:
float a = 60;
float b = 0.1;
float c = a + b * 10;
What is the value of c? It is not exactly 61, because b is not really .1 because .1 does not have an exact binary representation.
The number 61.0 does indeed have an exact floating-point operation—but that's not true for all integers. If you wrote a loop that added one to both a double-precision floating point number and a 64-bit integer, eventually you'd reach a point where the 64-bit integer perfectly represents a number, but the floating point doesn't—because there aren't enough significant bits.
It's just much easier to reach the point of approximation on the right side of the decimal point. If you started writing out all the numbers in binary floating point, it'd make more sense.
Another way of thinking about it is that when you note that 61.0 is perfectly representable in base 10, and shifting the decimal point around doesn't change that, you're performing multiplication by powers of ten (10^1, 10^-1). In floating point, multiplying by powers of two does not affect the precision of the number. Try taking 61.0 and dividing it by three repeatedly for an illustration of how a perfectly precise number can lose its precise representation.
There's a threshold because the meaning of the digit has gone from integer to non-integer. To represent 61, you have 6*10^1 + 1*10^0; 10^1 and 10^0 are both integers. 6.1 is 6*10^0 + 1*10^-1, but 10^-1 is 1/10, which is definitely not an integer. That's how you end up in Inexactville.
A parallel can be made of fractions and whole numbers. Some fractions eg 1/7 cannot be represented in decimal form without lots and lots of decimals. Because floating point is binary based the special cases change but the same sort of accuracy problems present themselves.
There are an infinite number of rational numbers, and a finite number of bits with which to represent them. See http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems.
you know integer numbers right? each bit represent 2^n
2^4=16
2^3=8
2^2=4
2^1=2
2^0=1
well its the same for floating point(with some distinctions) but the bits represent 2^-n
2^-1=1/2=0.5
2^-2=1/(2*2)=0.25
2^-3=0.125
2^-4=0.0625
Floating point binary representation:
sign Exponent Fraction(i think invisible 1 is appended to the fraction )
B11 B10 B9 B8 B7 B6 B5 B4 B3 B2 B1 B0
The high scoring answer above nailed it.
First you were mixing base 2 and base 10 in your question, then when you put a number on the right side that is not divisible into the base you get problems. Like 1/3 in decimal because 3 doesnt go into a power of 10 or 1/5 in binary which doesnt go into a power of 2.
Another comment though NEVER use equal with floating point numbers, period. Even if it is an exact representation there are some numbers in some floating point systems that can be accurately represented in more than one way (IEEE is bad about this, it is a horrible floating point spec to start with, so expect headaches). No different here 1/3 is not EQUAL to the number on your calculator 0.3333333, no matter how many 3's there are to the right of the decimal point. It is or can be close enough but is not equal. so you would expect something like 2*1/3 to not equal 2/3 depending on the rounding. Never use equal with floating point.
As we have been discussing, in floating point arithmetic, the decimal 0.1 cannot be perfectly represented in binary.
Floating point and integer representations provide grids or lattices for the numbers represented. As arithmetic is done, the results fall off the grid and have to be put back onto the grid by rounding. Example is 1/10 on a binary grid.
If we use binary coded decimal representation as one gentleman suggested, would we be able to keep numbers on the grid?
For a simple answer: The computer doesn't have infinite memory to store fraction (after representing the decimal number as the form of scientific notation). According to IEEE 754 standard for double-precision floating-point numbers, we only have a limit of 53 bits to store fraction.
For more info: http://mathcenter.oxford.emory.edu/site/cs170/ieee754/
I will not bother to repeat what the other 20 answers have already summarized, so I will just answer briefly:
The answer in your content:
Why can't base two numbers represent certain ratios exactly?
For the same reason that decimals are insufficient to represent certain ratios, namely, irreducible fractions with denominators containing prime factors other than two or five which will always have an indefinite string in at least the mantissa of its decimal expansion.
Why can't decimal numbers be represented exactly in binary?
This question at face value is based on a misconception regarding values themselves. No number system is sufficient to represent any quantity or ratio in a manner that the thing itself tells you that it is both a quantity, and at the same time also gives the interpretation in and of itself about the intrinsic value of the representation. As such, all quantitative representations, and models in general, are symbolic and can only be understood a posteriori, namely, after one has been taught how to read and interpret these numbers.
Since models are subjective things that are true insofar as they reflect reality, we do not strictly need to interpret a binary string as sums of negative and positive powers of two. Instead, one may observe that we can create an arbitrary set of symbols that use base two or any other base to represent any number or ratio exactly. Just consider that we can refer to all of infinity using a single word and even a single symbol without "showing infinity" itself.
As an example, I am designing a binary encoding for mixed numbers so that I can have more precision and accuracy than an IEEE 754 float. At the time of writing this, the idea is to have a sign bit, a reciprocal bit, a certain number of bits for a scalar to determine how much to "magnify" the fractional portion, and then the remaining bits are divided evenly between the integer portion of a mixed number, and the latter a fixed-point number which, if the reciprocal bit is set, should be interpreted as one divided by that number. This has the benefit of allowing me to represent numbers with infinite decimal expansions by using their reciprocals which do have terminating decimal expansions, or alternatively, as a fraction directly, potentially as an approximation, depending on my needs.
You can't represent 0.1 exactly in binary for the same reason you can't measure 0.1 inch using a conventional English ruler.
English rulers, like binary fractions, are all about halves. You can measure half an inch, or a quarter of an inch (which is of course half of a half), or an eighth, or a sixteenth, etc.
If you want to measure a tenth of an inch, though, you're out of luck. It's less than an eighth of an inch, but more than a sixteenth. If you try to get more exact, you find that it's a little more than 3/32, but a little less than 7/64. I've never seen an actual ruler that had gradations finer than 64ths, but if you do the math, you'll find that 1/10 is less than 13/128, and it's more than 25/256, and it's more than 51/512. You can keep going finer and finer, to 1024ths and 2048ths and 4096ths and 8192nds, but you will never find an exact marking, even on an infinitely-fine base-2 ruler, that exactly corresponds to 1/10, or 0.1.
You will find something interesting, though. Let's look at all the approximations I've listed, and for each one, record explicitly whether 0.1 is less or greater:
fraction
decimal
0.1 is...
as 0/1
1/2
0.5
less
0
1/4
0.25
less
0
1/8
0.125
less
0
1/16
0.0625
greater
1
3/32
0.09375
greater
1
7/64
0.109375
less
0
13/128
0.1015625
less
0
25/256
0.09765625
greater
1
51/512
0.099609375
greater
1
103/1024
0.1005859375
less
0
205/2048
0.10009765625
less
0
409/4096
0.099853515625
greater
1
819/8192
0.0999755859375
greater
1
Now, if you read down the last column, you get 0001100110011. It's no coincidence that the infinitely-repeating binary fraction for 1/10 is 0.0001100110011...
I want to perform some calculations and I want the result correct up to some decimal places, say 12.
So I wrote a sample:
#define PI 3.1415926535897932384626433832795028841971693993751
double d, k, h;
k = 999999/(2*PI);
h = 999999;
d = PI*k*k*h;
printf("%.12f\n", d);
But it gives the output:
79577232813771760.000000000000
I even used setprecision(), but same answer rather in exponential form.
cout<<setprecision(12)<<d<<endl;
prints
7.95772328138e+16
Used long double also, but in vain.
Now is there any way other than storing the integer part and the fractional part separately in long long int types?
If so, what can be done to get the answer precisely?
A double has only about 16 decimal digits of precision. Everything after the decimal point would be nonsense. (In fact, the last digit or two left of the point may not agree with an infinite-precision calculation.)
Long double is not standardized, AFAIK. It may be that on your system it is the same as double, or no more precise. That would slightly surprise me, but it doesn't violate anything.
You need to read Double-Precision concepts again; more carefully.
The double has increased precision by using 64 bits.
Stuff before the decimal is more important than that after it.
So, when you have a large integer part, it will truncate the lower precision -- this is being described to you in various answers here as rounding off.
Update:
To increase precision, you'll need to use some library or change your language.
Check this other question: Best coding language for dealing with large numbers (50000+ digits)
Yet, I'll ask you to re-check your intent once more.
Do you really need 12 decimal places for numbers that have really high values
(over 10 digits in the integer part like in your example)?
Maybe you won't really have large integer parts
(in which case such code should work fine).
But if you are tracking a value like 10000000000.123456789,
I am really interested in exactly which application you are working on (astronomy?).
If the integer part of your values is some way under 10000, you should be fine here.
Update2:
IF you must demonstrate the ability of a specific formula to work accurately within constrained error limits, the way to go is fixing the processing of your formula such that the least error is introduced.
Example,
If you want to do say, (x * y) / z
it would be prudent to try something like max(x,y)/z * min(x,y)
rather than, the original form which may overflow after (x * y), loosing precision if that did not fit in the 16 decimals of double
If you had just 2 digit precision,
. 2-digit regular-precision
`42 * 7 290 297
(42 * 7)/2 290/2 294/2
Result ==> 145 147
But ==> 42/2 = 21
21 * 7 = 147
This is probably the intent of your contest.
The double-precision binary format used by most computers can only hold about 16 digits, after that you'll get rounding. See http://en.wikipedia.org/wiki/Double-precision_floating-point_format
Floating point values have a limit range of digits. Just because your "PI" value has six times as many digits as a double will support doesn't alter the way the hardware works.
A typical (IEEE754) double will produce approximately 15-16 decimal places. Whether that's 0.12345678901235, 1234567.8901235, 12345678901235 or 12345678901235000000000, or some other variation.
In other words, yes, if you calculate your calculation EXACTLY, you'll get lots of decimal places, because pi never ends. On a computer, you get about 15-16 digits, no matter what input values you use - all that changes is where in that sequence the decimal place sits. To get more, you need "big number support", such as the Gnu Multiprcession (GMP) library.
You're looking for std::fixed. That tells the ostream not to use exponential form.
cout << setprecision(12) << std::fixed << d << endl;