How do you print the EXACT value of a floating point number?
First of all, this is not a floating point newbie question. I know results of floating point arithmetic (not to mention transcendental functions) usually cannot be represented exactly, and that most terminating decimals cannot be represented exactly as binary floating point numbers.
That said, each possible floating point value corresponds exactly to a diadic rational (a rational number p/q where q is a power of 2), which in turn has an exact decimal representation.
My question is: How do you find this exact decimal representation efficiently? sprintf and similar functions are usually only specified up to a number of significant digits to uniquely determine the original floating point value; they don't necessarily print the exact decimal representation. I know one algorithm 开发者_开发技巧I've used, but it's very slow, O(e^2) where e is the exponent. Here's an outline:
- Convert the mantissa to a decimal integer. You can either do this by pulling apart the bits to read the mantissa directly, or you can write a messy floating point loop that first multiplies the value by a power of two to put it in the range 1<=x<10, then pulls off a digit at a time by casting to int, subtracting, and multiplying by 10.
- Apply the exponent by repeatedly multiplying or dividing by 2. This is an operation on the string of decimal digits you generated. Every ~3 multiplications will add an extra digit to the left. Every single dividion will add an extra digit to the right.
Is this really the best possible? I doubt it, but I'm not a floating-point expert and I can't find a way to do the base-10 computations on the floating point representation of the number without running into a possibility of inexact results (multiplying or dividing by anything but a power of 2 is a lossy operation on floating point numbers unless you know you have free bits to work with).
This question has a bureaucratic part and an algorithmic part. A floating point number is stored internally as (2e × m), where e is an exponent (itself in binary) and m is a mantissa. The bureaucratic part of the question is how to access this data, but R. seems more interested in the algorithmic part of the question, namely, converting (2e × m) to a fraction (a/b) in decimal form. The answer to the bureaucratic question in several languages is frexp (which is an interesting detail that I didn’t know before today).
It is true that at first glance, it takes O(e2) work just to write 2e in decimal, and more time still for the mantissa. But, thanks to the magic of the Schönhage–Strassen fast multiplication algorithm, you can do it in Õ(e) time, where the tilde means “up to log factors”. If you view Schönhage–Strassen as magic, then it’s not that hard to think of what to do. If e is even, you can recursively compute 2e/2, and then square it using fast multiplication. On the other hand if e is odd, you can recursively compute 2e−1 and then double it. You have to be careful to check that there is a version of Schönhage–Strassen in base 10. Although it is not widely documented, it can be done in any base.
Converting a very long mantissa from binary to base 10 is not exactly the same question, but it has a similar answer. You can divide the mantissa into two halves, m = a × 2k + b. Then recursively convert a and b to base 10, convert 2k to base 10, and do another fast multiplication to compute m in base 10.
The abstract result behind all of this is that you can convert integers from one base to another in Õ(N) time.
If the question is about standard 64-bit floating point numbers, then it’s too small for the fancy Schönhage–Strassen algorithm. In this range you can instead save work with various tricks. One approach is to store all 2048 values of 2e in a lookup table, and then work in the mantissa with asymmetric multiplication (in between long multiplication and short multiplication). Another trick is to work in base 10000 (or a higher power of 10, depending on architecture) instead of base 10. But, as R. points out in the comments, 128-bit floating point numbers already allow large enough exponents to call into question both lookup tables and standard long multiplication. As a practical matter, long multiplication is the fastest up to a handful of digits, then in a significant medium range one can use Karatsuba multiplication or Toom–Cook multiplication, and then after that a variation of Schönhage–Strassen is best not just in theory but also in practice.
Actually, the big integer package GMP already has Õ(N)-time radix conversion, as well as good heuristics for which choice of multiplication algorithm. The only difference between their solution and mine is that instead of doing any big arithmetic in base 10, they compute large powers of 10 in base 2. In this solution, they also need fast division, but that can be obtained from fast multiplication in any of several ways.
I see you’ve accepted an answer already, but here are a couple of open source implementations of this conversion you might want to look at:
David Gay’s
dtoa()function indtoa.c: https://www.netlib.org/fp/dtoa.c.The function
___printf_fp()in the/stdio-common/printf_fp.cfile in Glibc (https://ftp.gnu.org/gnu/glibc/glibc-2.11.2.tar.gz, for example).
Both will print as many digits as you ask for in a %f-type printf, as I’ve written about at:
- https://www.exploringbinary.com/print-precision-of-dyadic-fractions-varies-by-language/ and
- https://www.exploringbinary.com/print-precision-of-floating-point-integers-varies-too/.
There's been a lot of work on printing floating-point numbers. The gold standard is to print out a decimal equivalent of minimal length such that when the decimal equivalent is read back in, you get the same floating-point number that you started with, no matter what the rounding mode is during readback. You can read about the algorithm in the excellent paper by Burger and Dybvig.
Although it's C# and your question is tagged with C, Jon Skeet has code to convert a double to its exact representation as a string: http://www.yoda.arachsys.com/csharp/DoubleConverter.cs
From a quick glance, it does not appear to be too hard to port to C, and even easier to write in C++.
Upon further reflection, it appears that Jon's algorithm is also O(e^2), as it also loops over the exponent. However, that means the algorithm is O(log(n)^2) (where n is the floating-point number), and I'm not sure you can convert from base 2 to base 10 in better than log-squared time.
Well being no floating point expert myself, I'd defer to using a well tested open source library.
The GNU MPFR is a good one.
The MPFR library is a C library for multiple-precision floating-point computations with correct rounding. The main goal of MPFR is to provide a library for multiple-precision floating-point computation which is both efficient and has a well-defined semantics.
sprintf and similar functions are usually only specified up to a number of significant digits to uniquely determine the original floating point value; they don't necessarily print the exact decimal representation.
You can ask for more significant digits than the default:
printf("%.100g\n", 0.1);
prints 0.1000000000000000055511151231257827021181583404541015625.
If you want more exact results, why not use fixed point math instead? Conversions are quick. Error is known and can be worked around. Not an exact answer to your question, but a different idea for you.
Off the top of my head, why not break the exponent down into a sum of binary exponents first, then all your operations are loss-less.
I.e.
10^2 = 2^6 + 2^5 + 2^2
Then sum:
mantissa<<6 + mantissa<<5 + mantissa<<2
I'm thinking that breaking it down would be on the O(n) on the the number of digits, the shifting is O(1), and the summing is O(n) digits...
You would have to have an integer class big enough to store the results, of course...
Let me know - I'm curious about this, it really made me think. :-)
There are 3 ways
printing numbers in
binorhexThis is the most precise way. I prefer
hexbecause it is more like base10for reading/feel like for exampleF.8h = 15.5no precision loss here.printing in
decbut only the relevant digitsWith this I mean only digits which can have
1in your number represented as close as possible.numof integer digits are easy and precise (no precision loss):// n10 - base 10 integer digits // n2 - base 2 integer digits n10=log10(2^n2) n10=log2(2^n2)/log2(10) n10=n2/log2(10) n10=ceil(n2*0.30102999566398119521373889472449) // if fist digit is 0 and n10 > 1 then n10--numof fractional digits are more tricky and empirically I found this:// n10 - base 10 fract. digits // n2 - base 2 fract. digits >= 8 n10=0; if (n02==8) n10=1; else if (n02==9) n10=2; else if (n02> 9) { n10=((n02-9)%10); if (n10>=6) n10=2; else if (n10>=1) n10=1; n10+=2+(((n02-9)/10)*3); }if you make a dependency table
n02 <-> n10then you see that constant0.30102999566398119521373889472449is still present, but at start from 8 bits because less cannot represent0.1with satisfactory precision (0.85 - 1.15). because of negative exponents of base2the dependency is not linear, instead it patterns. This code works for small bit count (<=52) but at larger bit counts there can be error because used pattern do not fitlog10(2)or1/log2(10)exactly.for larger bit counts I use this:
n10=7.810+(9.6366363636363636363636*((n02>>5)-1.0));but that formula is 32bit aligned !!! and also bigger bit count ads error to it
P.S. further analysis of binary representation of decadic numbers
0.1 0.01 0.001 0.0001 ...may reveal the exact pattern repetition which would lead to exact number of relevant digits for any bit count.
for clarity:
8 bin digits -> 1 dec digits 9 bin digits -> 2 dec digits 10 bin digits -> 3 dec digits 11 bin digits -> 3 dec digits 12 bin digits -> 3 dec digits 13 bin digits -> 3 dec digits 14 bin digits -> 3 dec digits 15 bin digits -> 4 dec digits 16 bin digits -> 4 dec digits 17 bin digits -> 4 dec digits 18 bin digits -> 4 dec digits 19 bin digits -> 5 dec digits 20 bin digits -> 6 dec digits 21 bin digits -> 6 dec digits 22 bin digits -> 6 dec digits 23 bin digits -> 6 dec digits 24 bin digits -> 6 dec digits 25 bin digits -> 7 dec digits 26 bin digits -> 7 dec digits 27 bin digits -> 7 dec digits 28 bin digits -> 7 dec digits 29 bin digits -> 8 dec digits 30 bin digits -> 9 dec digits 31 bin digits -> 9 dec digits 32 bin digits -> 9 dec digits 33 bin digits -> 9 dec digits 34 bin digits -> 9 dec digits 35 bin digits -> 10 dec digits 36 bin digits -> 10 dec digits 37 bin digits -> 10 dec digits 38 bin digits -> 10 dec digits 39 bin digits -> 11 dec digits 40 bin digits -> 12 dec digits 41 bin digits -> 12 dec digits 42 bin digits -> 12 dec digits 43 bin digits -> 12 dec digits 44 bin digits -> 12 dec digits 45 bin digits -> 13 dec digits 46 bin digits -> 13 dec digits 47 bin digits -> 13 dec digits 48 bin digits -> 13 dec digits 49 bin digits -> 14 dec digits 50 bin digits -> 15 dec digits 51 bin digits -> 15 dec digits 52 bin digits -> 15 dec digits 53 bin digits -> 15 dec digits 54 bin digits -> 15 dec digits 55 bin digits -> 16 dec digits 56 bin digits -> 16 dec digits 57 bin digits -> 16 dec digits 58 bin digits -> 16 dec digits 59 bin digits -> 17 dec digits 60 bin digits -> 18 dec digits 61 bin digits -> 18 dec digits 62 bin digits -> 18 dec digits 63 bin digits -> 18 dec digits 64 bin digits -> 18 dec digitsAnd at last do not forget to round the cut off digits !!! That means if digit after the last relevant digit is
>=5in dec than last relevant digit should be+1... and if it is already9than you must go to previous digit and so on...print exact value
To print exact value of fractional binary number just print fractional
ndigits wherenis the number of fractional bits because the value represented is sum of this values so the number of fractional decimals cannot be bigger than thenumof fractional digits of LSB. Stuff above (bullet #2) is relevant for storing decimal number tofloat(or printing just relevant decimals).negative powers of two exact values...
2^- 1 = 0.5 2^- 2 = 0.25 2^- 3 = 0.125 2^- 4 = 0.0625 2^- 5 = 0.03125 2^- 6 = 0.015625 2^- 7 = 0.0078125 2^- 8 = 0.00390625 2^- 9 = 0.001953125 2^-10 = 0.0009765625 2^-11 = 0.00048828125 2^-12 = 0.000244140625 2^-13 = 0.0001220703125 2^-14 = 0.00006103515625 2^-15 = 0.000030517578125 2^-16 = 0.0000152587890625 2^-17 = 0.00000762939453125 2^-18 = 0.000003814697265625 2^-19 = 0.0000019073486328125 2^-20 = 0.00000095367431640625now negative powers of
10printed by exact value style for 64 bitdoubles:10^+ -1 = 0.1000000000000000055511151231257827021181583404541015625 = 0.0001100110011001100110011001100110011001100110011001101b 10^+ -2 = 0.01000000000000000020816681711721685132943093776702880859375 = 0.00000010100011110101110000101000111101011100001010001111011b 10^+ -3 = 0.001000000000000000020816681711721685132943093776702880859375 = 0.000000000100000110001001001101110100101111000110101001111111b 10^+ -4 = 0.000100000000000000004792173602385929598312941379845142364501953125 = 0.000000000000011010001101101110001011101011000111000100001100101101b 10^+ -5 = 0.000010000000000000000818030539140313095458623138256371021270751953125 = 0.000000000000000010100111110001011010110001000111000110110100011110001b 10^+ -6 = 0.000000999999999999999954748111825886258685613938723690807819366455078125 = 0.000000000000000000010000110001101111011110100000101101011110110110001101b 10^+ -7 = 0.0000000999999999999999954748111825886258685613938723690807819366455078125 = 0.0000000000000000000000011010110101111111001010011010101111001010111101001b 10^+ -8 = 0.000000010000000000000000209225608301284726753266340892878361046314239501953125 = 0.000000000000000000000000001010101111001100011101110001000110000100011000011101b 10^+ -9 = 0.0000000010000000000000000622815914577798564188970686927859787829220294952392578125 = 0.0000000000000000000000000000010001001011100000101111101000001001101101011010010101b 10^+-10 = 0.00000000010000000000000000364321973154977415791655470655996396089904010295867919921875 = 0.00000000000000000000000000000000011011011111001101111111011001110101111011110110111011b 10^+-11 = 0.00000000000999999999999999939496969281939810930172340963650867706746794283390045166015625 = 0.00000000000000000000000000000000000010101111111010111111111100001011110010110010010010101b 10^+-12 = 0.00000000000099999999999999997988664762925561536725284350612952266601496376097202301025390625 = 0.00000000000000000000000000000000000000010001100101111001100110000001001011011110101000010001b 10^+-13 = 0.00000000000010000000000000000303737455634003709136034716842278413651001756079494953155517578125 = 0.00000000000000000000000000000000000000000001110000100101110000100110100001001001011101101000001b 10^+-14 = 0.000000000000009999999999999999988193093545598986971343290729163921781719182035885751247406005859375 = 0.000000000000000000000000000000000000000000000010110100001001001101110000110101000010010101110011011b 10^+-15 = 0.00000000000000100000000000000007770539987666107923830718560119501514549256171449087560176849365234375 = 0.00000000000000000000000000000000000000000000000001001000000011101011111001111011100111010101100001011b 10^+-16 = 0.00000000000000009999999999999999790977867240346035618411149408467364363417573258630000054836273193359375 = 0.00000000000000000000000000000000000000000000000000000111001101001010110010100101111101100010001001101111b 10^+-17 = 0.0000000000000000100000000000000007154242405462192450852805618492324772617063644020163337700068950653076171875 = 0.0000000000000000000000000000000000000000000000000000000010111000011101111010101000110010001101101010010010111b 10^+-18 = 0.00000000000000000100000000000000007154242405462192450852805618492324772617063644020163337700068950653076171875 = 0.00000000000000000000000000000000000000000000000000000000000100100111001001011101110100011101001001000011101011b 10^+-19 = 0.000000000000000000099999999999999997524592683526013185572915905567688179926555402943222361500374972820281982421875 = 0.000000000000000000000000000000000000000000000000000000000000000111011000001111001001010011111011011011010010101011b 10^+-20 = 0.00000000000000000000999999999999999945153271454209571651729503702787392447107715776066783064379706047475337982177734375 = 0.00000000000000000000000000000000000000000000000000000000000000000010111100111001010000100001100100100100100001000100011bnow negative powers of 10 printed by relevant decimal digits only style (i am more used to this) for 64bit
doubles:10^+ -1 = 0.1 10^+ -2 = 0.01 10^+ -3 = 0.001 10^+ -4 = 0.0001 10^+ -5 = 0.00001 10^+ -6 = 0.000001 10^+ -7 = 0.0000001 10^+ -8 = 0.00000001 10^+ -9 = 0.000000001 10^+-10 = 0.0000000001 10^+-11 = 0.00000000001 10^+-12 = 0.000000000001 10^+-13 = 0.0000000000001 10^+-14 = 0.00000000000001 10^+-15 = 0.000000000000001 10^+-16 = 0.0000000000000001 10^+-17 = 0.00000000000000001 10^+-18 = 0.000000000000000001 10^+-19 = 0.0000000000000000001 10^+-20 = 0.00000000000000000001hope it helps :)
You don't. The closest you can come to that is dumping the bytes.
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