I have seen long articles explaining how floating point numbers can be stored and how the arithmetic of those numbers is being done, but please briefly explain why when I write
cout << 1.0 / 3.0 <<endl;
I see 0.333333, but when I write
cout << 1.0 / 3.0 + 1.0 / 3.0 + 1.0 / 3.0 << endl;
I see 1.
How does the computer do this? Please explain just this simple example. It is enough for me.
Check out the article on "What every computer scientist should know about floating point arithmetic"
The problem is that the floating point format represents fractions in base 2.
The first fraction bit is ½, the second ¼, and it goes on as 1 / 2n.
And the problem with that is that not every rational number (a number that can be expressed as the ratio of two integers) actually has a finite representation in this base 2 format.
(This makes the floating point format difficult to use for monetary values. Although these values are always rational numbers (n/100) only .00, .25, .50, and .75 actually have exact representations in any number of digits of a base two fraction.
)
Anyway, when you add them back, the system eventually gets a chance to round the result to a number that it can represent exactly.
At some point, it finds itself adding the .666... number to the .333... one, like so:
00111110 1 .o10101010 10101010 10101011
+ 00111111 0 .10101010 10101010 10101011o
------------------------------------------
00111111 1 (1).0000000 00000000 0000000x # the x isn't in the final result
The leftmost bit is the sign, the next eight are the exponent, and the remaining bits are the fraction. In between the exponent and the fraction is an assummed "1" that is always present, and therefore not actually stored, as the normalized leftmost fraction bit. I've written zeroes that aren't actually present as individual bits as o.
A lot has happened here, at each step, the FPU has taken rather heroic measures to round the result. Two extra digits of precision (beyond what will fit in the result) have been kept, and the FPU knows in many cases if any, or at least 1, of the remaining rightmost bits were one. If so, then that part of the fraction is more than 0.5 (scaled) and so it rounds up. The intermediate rounded values allow the FPU to carry the rightmost bit all the way over to the integer part and finally round to the correct answer.
This didn't happen because anyone added 0.5; the FPU just did the best it could within the limitations of the format. Floating point is not, actually, inaccurate. It's perfectly accurate, but most of the numbers we expect to see in our base-10, rational-number world-view are not representable by the base-2 fraction of the format. In fact, very few are.
Let's do the math. For brevity, we assume that you only have four significant (base-2) digits.
Of course, since gcd(2,3)=1, 1/3 is periodic when represented in base-2. In particular, it cannot be represented exactly, so we need to content ourselves with the approximation
A := 1×1/4 + 0×1/8 + 1×1/16 + 1*1/32
which is closer to the real value of 1/3 than
A' := 1×1/4 + 0×1/8 + 1×1/16 + 0×1/32
So, printing A in decimal gives 0.34375 (the fact that you see 0.33333 in your example is just testament to the larger number of significant digits in a double).
When adding these up three times, we get
A + A + A
= ( A + A ) + A
= ( (1/4 + 1/16 + 1/32) + (1/4 + 1/16 + 1/32) ) + (1/4 + 1/16 + 1/32)
= ( 1/4 + 1/4 + 1/16 + 1/16 + 1/32 + 1/32 ) + (1/4 + 1/16 + 1/32)
= ( 1/2 + 1/8 + 1/16 ) + (1/4 + 1/16 + 1/32)
= 1/2 + 1/4 + 1/8 + 1/16 + 1/16 + O(1/32)
The O(1/32) term cannot be represented in the result, so it's discarded and we get
A + A + A = 1/2 + 1/4 + 1/8 + 1/16 + 1/16 = 1
QED :)
As for this specific example: I think the compilers are too clever nowadays, and automatically make sure a const result of primitive types will be exact if possible. I haven't managed to fool g++ into doing an easy calculation like this wrong.
However, it's easy to bypass such things by using non-const variables. Still,
int d = 3;
float a = 1./d;
std::cout << d*a;
will exactly yield 1, although this shouldn't really be expected. The reason, as was already said, is that the operator<< rounds the error away.
As to why it can do this: when you add numbers of similar size or multiply a float by an int, you get pretty much all the precision the float type can maximally offer you - that means, the ratio error/result is very small (in other words, the errors occur in a late decimal place, assuming you have a positive error).
So 3*(1./3), even though, as a float, not exactly ==1, has a big correct bias which prevents operator<< from taking care for the small errors. However, if you then remove this bias by just substracting 1, the floating point will slip down right to the error, and suddenly it's not neglectable at all any more. As I said, this doesn't happen if you just type 3*(1./3)-1 because the compiler is too clever, but try
int d = 3;
float a = 1./d;
std::cout << d*a << " - 1 = " << d*a - 1 << " ???\n";
What I get (g++, 32 bit Linux) is
1 - 1 = 2.98023e-08 ???
This works because the default precision is 6 digits, and rounded to 6 digits the result is 1. See 27.5.4.1 basic_ios constructors in the C++ draft standard (n3092).
Related
float x = 384.951257;
std::cout << std::fixed << std::setprecision(6) << x << std::endl;
The output is 384.951263. Why? I'm using gcc.
float is usually only 32-bit. With about 3 bits per decimal digit (210 roughly equals 103) that means it can't possibly represent more than about 11 decimal digits, and accounting for other information it also needs to represent, such as magnitude, let's say 6-7 decimal digits. Hey, that's what you got!
Check e.g. Wikipedia for details.
Use double or long double for better precision. double is the default in C++. E.g., the literal 3.14 is of type double.
Floats have a limited resolution. So it gets rounded when you assing the value to x.
All answers here talk as though the issue is due to floating-point numbers and their capacity, but those are just implementation details; the issue is deeper than that. This issue occurs when representing decimal numbers using binary number system. Even something as simple as 0.1)10 is not precisely representable in binary, since it can only represent those numbers as a finite fraction where the denominator is a power of 2. Unfortunately, this does not include most of the numbers that can be represented as finite fraction in base 10, like 0.1.
The single-precision float datatype usually gets mapped to binary32 as called by the IEEE 754 standard, has 32-bits which is partitioned into 1 sign bit, 8 exponent bits and 23 significand bits (excluding the hidden/implicit bit). Thus we've to calculate upto 24 bits when converting to binary32.
Other answers here evade the actual calculations involved, I'll try to do it. This method is explained in greater detail here. So lets convert the real number into a binary number:
Integer part 384)10 = 110000000)2 (using the usual method of successive division by 2)
Fractional part 0.951257)10 can be converted by successive multiplication by 2 and taking the integer part
0.951257 * 2 = 1.902514
0.902514 * 2 = 1.805028
0.805028 * 2 = 1.610056
0.610056 * 2 = 1.220112
0.220112 * 2 = 0.440224
0.440224 * 2 = 0.880448
0.880448 * 2 = 1.760896
0.760896 * 2 = 1.521792
0.521792 * 2 = 1.043584
0.043584 * 2 = 0.087168
0.087168 * 2 = 0.174336
0.174336 * 2 = 0.348672
0.348672 * 2 = 0.697344
0.697344 * 2 = 1.394688
0.394688 * 2 = 0.789376
Gathering the obtined fractional part in binary we've 0.111100111000010)2. The overall number in binary would be 110000000.111100111000010)2; this has 24 bits as required.
Converting this back to decimal would give you 384 + (15585 / 16384) = 384.951232)10. With the rounding mode (round to nearest) enabled this comes to, what you see, 384.951263)10.
This can be verified here.
I had a problem when I was adding three floating point values and comparing them to 1.
cout << ((0.7 + 0.2 + 0.1)==1)<<endl; //output is 0
cout << ((0.7 + 0.1 + 0.2)==1)<<endl; //output is 1
Why would these values come out different?
Floating point addition is not necessarily associative. If you change the order in which you add things up, this can change the result.
The standard paper on the subject is What Every Computer Scientist Should Know about Floating Point Arithmetic. It gives the following example:
Another grey area concerns the interpretation of parentheses. Due to roundoff errors, the associative laws of algebra do not necessarily hold for floating-point numbers. For example, the expression (x+y)+z has a totally different answer than x+(y+z) when x = 1e30, y = -1e30 and z = 1 (it is 1 in the former case, 0 in the latter).
What is likely, with currently popular machines and software, is:
The compiler encoded .7 as 0x1.6666666666666p-1 (this is the hexadecimal numeral 1.6666666666666 multiplied by 2 to the power of -1), .2 as 0x1.999999999999ap-3, and .1 as 0x1.999999999999ap-4. Each of these is the number representable in floating-point that is closest to the decimal numeral you wrote.
Observe that each of these hexadecimal floating-point constants has exactly 53 bits in its significand (the "fraction" part, often inaccurately called the mantissa). The hexadecimal numeral for the significand has a "1" and thirteen more hexadecimal digits (four bits each, 52 total, 53 including the "1"), which is what the IEEE-754 standard provides for, for 64-bit binary floating-point numbers.
Let's add the numbers for .7 and .2: 0x1.6666666666666p-1 and 0x1.999999999999ap-3. First, scale the exponent of the second number to match the first. To do this, we will multiply the exponent by 4 (changing "p-3" to "p-1") and multiply the significand by 1/4, giving 0x0.66666666666668p-1. Then add 0x1.6666666666666p-1 and 0x0.66666666666668p-1, giving 0x1.ccccccccccccc8p-1. Note that this number has more than 53 bits in the significand: The "8" is the 14th digit after the period. Floating-point cannot return a result with this many bits, so it has to be rounded to the nearest representable number. In this case, there are two numbers that are equally near, 0x1.cccccccccccccp-1 and 0x1.ccccccccccccdp-1. When there is a tie, the number with a zero in the lowest bit of the significand is used. "c" is even and "d" is odd, so "c" is used. The final result of the addition is 0x1.cccccccccccccp-1.
Next, add the number for .1 (0x1.999999999999ap-4) to that. Again, we scale to make the exponents match, so 0x1.999999999999ap-4 becomes 0x.33333333333334p-1. Then add that to 0x1.cccccccccccccp-1, giving 0x1.fffffffffffff4p-1. Rounding that to 53 bits gives 0x1.fffffffffffffp-1, and that is the final result of .7+.2+.1.
Now consider .7+.1+.2. For .7+.1, add 0x1.6666666666666p-1 and 0x1.999999999999ap-4. Recall the latter is scaled to 0x.33333333333334p-1. Then the exact sum is 0x1.99999999999994p-1. Rounding that to 53 bits gives 0x1.9999999999999p-1.
Then add the number for .2 (0x1.999999999999ap-3), which is scaled to 0x0.66666666666668p-1. The exact sum is 0x2.00000000000008p-1. Floating-point significands are always scaled to start with 1 (except for special cases: zero, infinity, and very small numbers at the bottom of the representable range), so we adjust this to 0x1.00000000000004p0. Finally, we round to 53 bits, giving 0x1.0000000000000p0.
Thus, because of errors that occur when rounding, .7+.2+.1 returns 0x1.fffffffffffffp-1 (very slightly less than 1), and .7+.1+.2 returns 0x1.0000000000000p0 (exactly 1).
Floating point multiplication is not associative in C or C++.
Proof:
#include<stdio.h>
#include<time.h>
#include<stdlib.h>
using namespace std;
int main() {
int counter = 0;
srand(time(NULL));
while(counter++ < 10){
float a = rand() / 100000;
float b = rand() / 100000;
float c = rand() / 100000;
if (a*(b*c) != (a*b)*c){
printf("Not equal\n");
}
}
printf("DONE");
return 0;
}
In this program, about 30% of the time, (a*b)*c is not equal to a*(b*c).
Neither addition nor multiplication is associative with IEEE 743 double precision (64-bit) numbers. Here are examples for each (evaluated with Python 3.9.7):
>>> (.1 + .2) + .3
0.6000000000000001
>>> .1 + (.2 + .3)
0.6
>>> (.1 * .2) * .3
0.006000000000000001
>>> .1 * (.2 * .3)
0.006
Similar answer to Eric's, but for addition, and with Python.
import random
random.seed(0)
n = 1000
a = [random.random() for i in range(n)]
b = [random.random() for i in range(n)]
c = [random.random() for i in range(n)]
sum(1 if (a[i] + b[i]) + c[i] != a[i] + (b[i] + c[i]) else 0 for i in range(n))
I've read a few texts and threads showing how to convert from a decimal to IEEE 754 but I am still confused as to how I can convert the number without expanding the decimal (which is represented in scientific notation)
The number I am particularly working with is 9.07 * 10^23, but any number would do; I will figure out how to do it for my particular example.
I'm assuming you want the result to be the floating-point number closest to the decimal number, and that you are using double-precision floating-point numbers.
For most numbers, there is a way to do it relatively quickly. Here's how it works in a nutshell.
You need to split the number into either a product or a fraction of numbers that have an exact representation as a floating-point number. The largest power of 10 that is exactly representable is 10^22. So, to get 9.07e+23 in floating-point form, we can write:
9.07e+23 = 907 * 10^21
According to the IEEE-754 standard, a single floating-point operation is guaranteed to be correctly rounded, so the above product, computed as a product of 2 double precision floating-point numbers, will give the correctly rounded result.
If you were to use this in a conversion function, you would probably store the powers of 10 in an array.
Note that you can't use this method for 9.07e-23. This number equals 907 / 10^23, so the denominator would be too large to be exactly representable. In this situation, and other dealings with very large or very small numbers, you have to use some form of high-precision arithmetic.
See Fast Path Decimal to Floating-Point Conversion for further details and examples.
Converting a number from a decimal string to binary IEEE is fairly straight-forward if you know how to do IEEE floating-point addition and multiplication. (or if you're using any basic programming language like C/C++)
There's a lot of different approaches to this, but the easiest is to evaluate 9.07 * 10^23 directly.
First, start with 9.07:
9.07 = 9 + 0 * 10^-1 + 7 * 10^-2
Now evaluate 10^23. This can be done by starting with 10 and using any powering algorithm.
Then multiply the results together.
Here's a simple implementation in C/C++:
double mantissa = 9;
mantissa += 0 / 10.;
mantissa += 7 / 100.;
double exp = 1;
for (int i = 0; i < 23; i++){
exp *= 10;
}
double result = mantissa * exp;
Now, going backwards (IEEE -> to decimal) is a lot harder.
Again, there's also a lot of different approaches. Here's the easiest one I can think of it.
I'll use 1.0011101b * 2^40 as the example. (the mantissa is in binary)
First, convert the mantissa to decimal: (this should be easy, since there's no exponent)
1.0011101b * 2^40 = 1.22656 * 2^40
Now, "scale" the number such that the binary exponent vanishes. This is done by multiplying by an appropriate power of 10 to "get rid" of the binary exponent.
1.22656 * 2^40 = 1.22656 * (2^40 * 10^-12) * 10^12
= 1.22656 * (1.09951) * 10^12
= 1.34861 * 10^12
So the answer is:
1.0011101b * 2^40 = 1.34861 * 10^12
In this example, 10^12 was needed to "scale away" the 2^40. Determining the power of 10 that is needed is simply equal to:
power of 10 = (power of 2) * log(2)/log(10)
How do you print a double to a stream so that when it is read in you don't lose precision?
I tried:
std::stringstream ss;
double v = 0.1 * 0.1;
ss << std::setprecision(std::numeric_limits<T>::digits10) << v << " ";
double u;
ss >> u;
std::cout << "precision " << ((u == v) ? "retained" : "lost") << std::endl;
This did not work as I expected.
But I can increase precision (which surprised me as I thought that digits10 was the maximum required).
ss << std::setprecision(std::numeric_limits<T>::digits10 + 2) << v << " ";
// ^^^^^^ +2
It has to do with the number of significant digits and the first two don't count in (0.01).
So has anybody looked at representing floating point numbers exactly?
What is the exact magical incantation on the stream I need to do?
After some experimentation:
The trouble was with my original version. There were non-significant digits in the string after the decimal point that affected the accuracy.
So to compensate for this we can use scientific notation to compensate:
ss << std::scientific
<< std::setprecision(std::numeric_limits<double>::digits10 + 1)
<< v;
This still does not explain the need for the +1 though.
Also if I print out the number with more precision I get more precision printed out!
std::cout << std::scientific << std::setprecision(std::numeric_limits<double>::digits10) << v << "\n";
std::cout << std::scientific << std::setprecision(std::numeric_limits<double>::digits10 + 1) << v << "\n";
std::cout << std::scientific << std::setprecision(std::numeric_limits<double>::digits) << v << "\n";
It results in:
1.000000000000000e-02
1.0000000000000002e-02
1.00000000000000019428902930940239457413554200000000000e-02
Based on #Stephen Canon answer below:
We can print out exactly by using the printf() formatter, "%a" or "%A". To achieve this in C++ we need to use the fixed and scientific manipulators (see n3225: 22.4.2.2.2p5 Table 88)
std::cout.flags(std::ios_base::fixed | std::ios_base::scientific);
std::cout << v;
For now I have defined:
template<typename T>
std::ostream& precise(std::ostream& stream)
{
std::cout.flags(std::ios_base::fixed | std::ios_base::scientific);
return stream;
}
std::ostream& preciselngd(std::ostream& stream){ return precise<long double>(stream);}
std::ostream& precisedbl(std::ostream& stream) { return precise<double>(stream);}
std::ostream& preciseflt(std::ostream& stream) { return precise<float>(stream);}
Next: How do we handle NaN/Inf?
It's not correct to say "floating point is inaccurate", although I admit that's a useful simplification. If we used base 8 or 16 in real life then people around here would be saying "base 10 decimal fraction packages are inaccurate, why did anyone ever cook those up?".
The problem is that integral values translate exactly from one base into another, but fractional values do not, because they represent fractions of the integral step and only a few of them are used.
Floating point arithmetic is technically perfectly accurate. Every calculation has one and only one possible result. There is a problem, and it is that most decimal fractions have base-2 representations that repeat. In fact, in the sequence 0.01, 0.02, ... 0.99, only a mere 3 values have exact binary representations. (0.25, 0.50, and 0.75.) There are 96 values that repeat and therefore are obviously not represented exactly.
Now, there are a number of ways to write and read back floating point numbers without losing a single bit. The idea is to avoid trying to express the binary number with a base 10 fraction.
Write them as binary. These days, everyone implements the IEEE-754 format so as long as you choose a byte order and write or read only that byte order, then the numbers will be portable.
Write them as 64-bit integer values. Here you can use the usual base 10. (Because you are representing the 64-bit aliased integer, not the 52-bit fraction.)
You can also just write more decimal fraction digits. Whether this is bit-for-bit accurate will depend on the quality of the conversion libraries and I'm not sure I would count on perfect accuracy (from the software) here. But any errors will be exceedingly small and your original data certainly has no information in the low bits. (None of the constants of physics and chemistry are known to 52 bits, nor has any distance on earth ever been measured to 52 bits of precision.) But for a backup or restore where bit-for-bit accuracy might be compared automatically, this obviously isn't ideal.
Don't print floating-point values in decimal if you don't want to lose precision. Even if you print enough digits to represent the number exactly, not all implementations have correctly-rounded conversions to/from decimal strings over the entire floating-point range, so you may still lose precision.
Use hexadecimal floating point instead. In C:
printf("%a\n", yourNumber);
C++0x provides the hexfloat manipulator for iostreams that does the same thing (on some platforms, using the std::hex modifier has the same result, but this is not a portable assumption).
Using hex floating point is preferred for several reasons.
First, the printed value is always exact. No rounding occurs in writing or reading a value formatted in this way. Beyond the accuracy benefits, this means that reading and writing such values can be faster with a well tuned I/O library. They also require fewer digits to represent values exactly.
I got interested in this question because I'm trying to (de)serialize my data to & from JSON.
I think I have a clearer explanation (with less hand waiving) for why 17 decimal digits are sufficient to reconstruct the original number losslessly:
Imagine 3 number lines:
1. for the original base 2 number
2. for the rounded base 10 representation
3. for the reconstructed number (same as #1 because both in base 2)
When you convert to base 10, graphically, you choose the tic on the 2nd number line closest to the tic on the 1st. Likewise when you reconstruct the original from the rounded base 10 value.
The critical observation I had was that in order to allow exact reconstruction, the base 10 step size (quantum) has to be < the base 2 quantum. Otherwise, you inevitably get the bad reconstruction shown in red.
Take the specific case of when the exponent is 0 for the base2 representation. Then the base2 quantum will be 2^-52 ~= 2.22 * 10^-16. The closest base 10 quantum that's less than this is 10^-16. Now that we know the required base 10 quantum, how many digits will be needed to encode all possible values? Given that we're only considering the case of exponent = 0, the dynamic range of values we need to represent is [1.0, 2.0). Therefore, 17 digits would be required (16 digits for fraction and 1 digit for integer part).
For exponents other than 0, we can use the same logic:
exponent base2 quant. base10 quant. dynamic range digits needed
---------------------------------------------------------------------
1 2^-51 10^-16 [2, 4) 17
2 2^-50 10^-16 [4, 8) 17
3 2^-49 10^-15 [8, 16) 17
...
32 2^-20 10^-7 [2^32, 2^33) 17
1022 9.98e291 1.0e291 [4.49e307,8.99e307) 17
While not exhaustive, the table shows the trend that 17 digits are sufficient.
Hope you like my explanation.
In C++20 you'll be able to use std::format to do this:
std::stringstream ss;
double v = 0.1 * 0.1;
ss << std::format("{}", v);
double u;
ss >> u;
assert(v == u);
The default floating-point format is the shortest decimal representation with a round-trip guarantee. The advantage of this method compared to using the precision of max_digits10 (not digits10 which is not suitable for round trip through decimal) from std::numeric_limits is that it doesn't print unnecessary digits.
In the meantime you can use the {fmt} library, std::format is based on. For example (godbolt):
fmt::print("{}", 0.1 * 0.1);
Output (assuming IEEE754 double):
0.010000000000000002
{fmt} uses the Dragonbox algorithm for fast binary floating point to decimal conversion. In addition to giving the shortest representation it is 20-30x faster than common standard library implementations of printf and iostreams.
Disclaimer: I'm the author of {fmt} and C++20 std::format.
A double has the precision of 52 binary digits or 15.95 decimal digits. See http://en.wikipedia.org/wiki/IEEE_754-2008. You need at least 16 decimal digits to record the full precision of a double in all cases. [But see fourth edit, below].
By the way, this means significant digits.
Answer to OP edits:
Your floating point to decimal string runtime is outputing way more digits than are significant. A double can only hold 52 bits of significand (actually, 53, if you count a "hidden" 1 that is not stored). That means the the resolution is not more than 2 ^ -53 = 1.11e-16.
For example: 1 + 2 ^ -52 = 1.0000000000000002220446049250313 . . . .
Those decimal digits, .0000000000000002220446049250313 . . . . are the smallest binary "step" in a double when converted to decimal.
The "step" inside the double is:
.0000000000000000000000000000000000000000000000000001 in binary.
Note that the binary step is exact, while the decimal step is inexact.
Hence the decimal representation above,
1.0000000000000002220446049250313 . . .
is an inexact representation of the exact binary number:
1.0000000000000000000000000000000000000000000000000001.
Third Edit:
The next possible value for a double, which in exact binary is:
1.0000000000000000000000000000000000000000000000000010
converts inexactly in decimal to
1.0000000000000004440892098500626 . . . .
So all of those extra digits in the decimal are not really significant, they are just base conversion artifacts.
Fourth Edit:
Though a double stores at most 16 significant decimal digits, sometimes 17 decimal digits are necessary to represent the number. The reason has to do with digit slicing.
As I mentioned above, there are 52 + 1 binary digits in the double. The "+ 1" is an assumed leading 1, and is neither stored nor significant. In the case of an integer, those 52 binary digits form a number between 0 and 2^53 - 1. How many decimal digits are necessary to store such a number? Well, log_10 (2^53 - 1) is about 15.95. So at most 16 decimal digits are necessary. Let's label these d_0 to d_15.
Now consider that IEEE floating point numbers also have an binary exponent. What happens when we increment the exponet by, say, 2? We have multiplied our 52-bit number, whatever it was, by 4. Now, instead of our 52 binary digits aligning perfectly with our decimal digits d_0 to d_15, we have some significant binary digits represented in d_16. However, since we multiplied by something less than 10, we still have significant binary digits represented in d_0. So our 15.95 decimal digits now occuply d_1 to d_15, plus some upper bits of d_0 and some lower bits of d_16. This is why 17 decimal digits are sometimes needed to represent a IEEE double.
Fifth Edit
Fixed numerical errors
The easiest way (for IEEE 754 double) to guarantee a round-trip conversion is to always use 17 significant digits. But that has the disadvantage of sometimes including unnecessary noise digits (0.1 → "0.10000000000000001").
An approach that's worked for me is to sprintf the number with 15 digits of precision, then check if atof gives you back the original value. If it doesn't, try 16 digits. If that doesn't work, use 17.
You might want to try David Gay's algorithm (used in Python 3.1 to implement float.__repr__).
Thanks to ThomasMcLeod for pointing out the error in my table computation
To guarantee round-trip conversion using 15 or 16 or 17 digits is only possible for a comparatively few cases. The number 15.95 comes from taking 2^53 (1 implicit bit + 52 bits in the significand/"mantissa") which comes out to an integer in the range 10^15 to 10^16 (closer to 10^16).
Consider a double precision value x with an exponent of 0, i.e. it falls into the floating point range range 1.0 <= x < 2.0. The implicit bit will mark the 2^0 component (part) of x. The highest explicit bit of the significand will denote the next lower exponent (from 0) <=> -1 => 2^-1 or the 0.5 component.
The next bit 0.25, the ones after 0.125, 0.0625, 0.03125, 0.015625 and so on (see table below). The value 1.5 will thus be represented by two components added together: the implicit bit denoting 1.0 and the highest explicit significand bit denoting 0.5.
This illustrates that from the implicit bit downward you have 52 additional, explicit bits to represent possible components where the smallest is 0 (exponent) - 52 (explicit bits in significand) = -52 => 2^-52 which according to the table below is ... well you can see for yourselves that it comes out to quite a bit more than 15.95 significant digits (37 to be exact). To put it another way the smallest number in the 2^0 range that is != 1.0 itself is 2^0+2^-52 which is 1.0 + the number next to 2^-52 (below) = (exactly) 1.0000000000000002220446049250313080847263336181640625, a value which I count as being 53 significant digits long. With 17 digit formatting "precision" the number will display as 1.0000000000000002 and this would depend on the library converting correctly.
So maybe "round-trip conversion in 17 digits" is not really a concept that is valid (enough).
2^ -1 = 0.5000000000000000000000000000000000000000000000000000
2^ -2 = 0.2500000000000000000000000000000000000000000000000000
2^ -3 = 0.1250000000000000000000000000000000000000000000000000
2^ -4 = 0.0625000000000000000000000000000000000000000000000000
2^ -5 = 0.0312500000000000000000000000000000000000000000000000
2^ -6 = 0.0156250000000000000000000000000000000000000000000000
2^ -7 = 0.0078125000000000000000000000000000000000000000000000
2^ -8 = 0.0039062500000000000000000000000000000000000000000000
2^ -9 = 0.0019531250000000000000000000000000000000000000000000
2^-10 = 0.0009765625000000000000000000000000000000000000000000
2^-11 = 0.0004882812500000000000000000000000000000000000000000
2^-12 = 0.0002441406250000000000000000000000000000000000000000
2^-13 = 0.0001220703125000000000000000000000000000000000000000
2^-14 = 0.0000610351562500000000000000000000000000000000000000
2^-15 = 0.0000305175781250000000000000000000000000000000000000
2^-16 = 0.0000152587890625000000000000000000000000000000000000
2^-17 = 0.0000076293945312500000000000000000000000000000000000
2^-18 = 0.0000038146972656250000000000000000000000000000000000
2^-19 = 0.0000019073486328125000000000000000000000000000000000
2^-20 = 0.0000009536743164062500000000000000000000000000000000
2^-21 = 0.0000004768371582031250000000000000000000000000000000
2^-22 = 0.0000002384185791015625000000000000000000000000000000
2^-23 = 0.0000001192092895507812500000000000000000000000000000
2^-24 = 0.0000000596046447753906250000000000000000000000000000
2^-25 = 0.0000000298023223876953125000000000000000000000000000
2^-26 = 0.0000000149011611938476562500000000000000000000000000
2^-27 = 0.0000000074505805969238281250000000000000000000000000
2^-28 = 0.0000000037252902984619140625000000000000000000000000
2^-29 = 0.0000000018626451492309570312500000000000000000000000
2^-30 = 0.0000000009313225746154785156250000000000000000000000
2^-31 = 0.0000000004656612873077392578125000000000000000000000
2^-32 = 0.0000000002328306436538696289062500000000000000000000
2^-33 = 0.0000000001164153218269348144531250000000000000000000
2^-34 = 0.0000000000582076609134674072265625000000000000000000
2^-35 = 0.0000000000291038304567337036132812500000000000000000
2^-36 = 0.0000000000145519152283668518066406250000000000000000
2^-37 = 0.0000000000072759576141834259033203125000000000000000
2^-38 = 0.0000000000036379788070917129516601562500000000000000
2^-39 = 0.0000000000018189894035458564758300781250000000000000
2^-40 = 0.0000000000009094947017729282379150390625000000000000
2^-41 = 0.0000000000004547473508864641189575195312500000000000
2^-42 = 0.0000000000002273736754432320594787597656250000000000
2^-43 = 0.0000000000001136868377216160297393798828125000000000
2^-44 = 0.0000000000000568434188608080148696899414062500000000
2^-45 = 0.0000000000000284217094304040074348449707031250000000
2^-46 = 0.0000000000000142108547152020037174224853515625000000
2^-47 = 0.0000000000000071054273576010018587112426757812500000
2^-48 = 0.0000000000000035527136788005009293556213378906250000
2^-49 = 0.0000000000000017763568394002504646778106689453125000
2^-50 = 0.0000000000000008881784197001252323389053344726562500
2^-51 = 0.0000000000000004440892098500626161694526672363281250
2^-52 = 0.0000000000000002220446049250313080847263336181640625
#ThomasMcLeod: I think the significant digit rule comes from my field, physics, and means something more subtle:
If you have a measurement that gets you the value 1.52 and you cannot read any more detail off the scale, and say you are supposed to add another number (for example of another measurement because this one's scale was too small) to it, say 2, then the result (obviously) has only 2 decimal places, i.e. 3.52.
But likewise, if you add 1.1111111111 to the value 1.52, you get the value 2.63 (and nothing more!).
The reason for the rule is to prevent you from kidding yourself into thinking you got more information out of a calculation than you put in by the measurement (which is impossible, but would seem that way by filling it with garbage, see above).
That said, this specific rule is for addition only (for addition: the error of the result is the sum of the two errors - so if you measure just one badly, though luck, there goes your precision...).
How to get the other rules:
Let's say a is the measured number and δa the error. Let's say your original formula was:
f:=m a
Let's say you also measure m with error δm (let that be the positive side).
Then the actual limit is:
f_up=(m+δm) (a+δa)
and
f_down=(m-δm) (a-δa)
So,
f_up =m a+δm δa+(δm a+m δa)
f_down=m a+δm δa-(δm a+m δa)
Hence, now the significant digits are even less:
f_up ~m a+(δm a+m δa)
f_down~m a-(δm a+m δa)
and so
δf=δm a+m δa
If you look at the relative error, you get:
δf/f=δm/m+δa/a
For division it is
δf/f=δm/m-δa/a
Hope that gets the gist across and hope I didn't make too many mistakes, it's late here :-)
tl,dr: Significant digits mean how many of the digits in the output actually come from the digits in your input (in the real world, not the distorted picture that floating point numbers have).
If your measurements were 1 with "no" error and 3 with "no" error and the function is supposed to be 1/3, then yes, all infinite digits are actual significant digits. Otherwise, the inverse operation would not work, so obviously they have to be.
If significant digit rule means something completely different in another field, carry on :-)
I am writing a piece of code in which i have to convert from double to float values. I am using boost::numeric_cast to do this conversion which will alert me of any overflow/underflow. However i am also interested in knowing if that conversion resulted in some precision loss or not.
For example
double source = 1988.1012;
float dest = numeric_cast<float>(source);
Produces dest which has value 1988.1
Is there any way available in which i can detect this kind of precision loss/rounding
You could cast the float back to a double and compare this double to the original - that should give you a fair indication as to whether there was a loss of precision.
float dest = numeric_cast<float>(source);
double residual = source - numeric_cast<double>(dest);
Hence, residual contains the "loss" you're looking for.
Look at these articles for single precision and double precision floats. First of all, floats have 8 bits for the exponent vs. 11 for a double. So anything bigger than 10^127 or smaller than 10^-126 in magnitude is going to be the overflow as you mentioned. For the float, you have 23 bits for the actual digits of the number, vs 52 bits for the double. So obviously, you have a lot more digits of precision for the double than float.
Say you have a number like: 1.1123. This number may not actually be encoded as 1.1123 because the digits in a floating point number are used to actually add up as fractions. For example, if your bits in the mantissa were 11001, then the value would be formed by 1 (implicit) + 1 * 1/2 + 1 * 1/4 + 0 * 1/8 + 0 * 1/16 + 1 * 1/32 + 0 * (64 + 128 + ...). So the exact value cannot be encoded unless you can add up these fractions in such a way that it's the exact number. This is rare. Therefore, there will almost always be a precision loss.
You're going to have a certain level of precision loss, as per Dave's answer. If, however, you want to focus on quantifying it and raising an exception when it exceeds a certain number, you will have to open up the floating point number itself and parse out the mantissa & exponent, then do some analysis to determine if you've exceeded your tolerance.
But, the good news, its usually the standard IEEE floating-point float. :-)