Formatting doubles for output in C

Question

Running a quick experiment related to Is double Multiplication Broken in  NET  and reading a couple of articles on C  string formatting  I thought that this         double i   10   0 69      Console WriteLine i       Console WriteLine String Format     0 F20    i        Console WriteLine String Format     0 F20    6 9 - i        Console WriteLine String Format     0 F20    6 9        Would be the C  equivalent of this C code         double i   10   0 69       printf     f n   i        printf        20f n   i        printf        20f n   6 9 - i        printf        20f n   6 9        However the C  produces the output   6 9   6 90000000000000000000   0 00000000000000088818   6 90000000000000000000   despite i showing up equal to the value 6 89999999999999946709  rather than 6 9  in the debugger   compared with C which shows the precision requested by the format   6 900000                             6 89999999999999946709             0 00000000000000088818             6 90000000000000035527             What s going on     Microsoft  NET Framework Version 3 51 SP1   Visual Studio C  2008 Express Edition      I have a background in numerical computing and experience implementing interval arithmetic - a technique for estimating errors due to the limits of precision in complicated numerical systems - on various platforms  To get the bounty  don t try and explain about the storage precision - in this case it s a difference of one ULP of a 64 bit double    To get the bounty  I want to know how  or whether   Net can format a double to the requested precision as visible in the C code

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I found this quick fix       double i   10   0 69      System Diagnostics Debug WriteLine i         String s   String Format   0 F20    i  Substring 0 20       System Diagnostics Debug WriteLine s        s Length

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Use  Console WriteLine String Format     0 G17    i      That will give you all the 17 digits it have  By default  a Double value contains 15 decimal digits of precision  although a maximum of 17 digits is maintained internally   0 R  will not always give you 17 digits  it will give 15 if the number can be represented with that precision   which returns 15 digits if the number can be represented with that precision or 17 digits if the number can only be represented with maximum precision  There isn t any thing you can to do to make the the double return more digits that is the way it s implemented  If you don t like it do a new double class yourself      NET s double cant store any more digits than 17 so you cant see 6 89999999999999946709 in the debugger you would see 6 8999999999999995  Please provide an image to prove us wrong

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Though this question is meanwhile closed  I believe it is worth mentioning how this atrocity came into existence  In a way  you may blame the C  spec  which states that a double must have a precision of 15 or 16 digits  the result of IEEE-754   A bit further on  section 4 1 6  it s stated that implementations are allowed to use higher precision  Mind you  higher  not lower  They are even allowed to deviate from IEEE-754  expressions of the type x   y   z where x   y would yield   -INF but would be in a valid range after dividing  do not have to result in an error  This feature makes it easier for compilers to use higher precision in architectures where that d yield better performance   But I promised a  reason   Here s a quote  you requested a resource in one of your recent comments  from the Shared Source CLI  in clr src vm comnumber cpp       In order to give numbers that are both   friendly to display and   round-trippable  we parse the number   using 15 digits and then determine if   it round trips to the same value   If   it does  we convert that NUMBER to a   string  otherwise we reparse using 17   digits and display that     In other words  MS s CLI Development Team decided to be both round-trippable and show pretty values that aren t such a pain to read  Good or bad  I d wish for an opt-in or opt-out   The trick it does to find out this round-trippability of any given number  Conversion to a generic NUMBER structure  which has separate fields for the properties of a double  and back  and then compare whether the result is different  If it is different  the exact value is used  as in your middle value with 6 9 - i  if it is the same  the  pretty value  is used    As you already remarked in a comment to Andyp  6 90   00 is bitwise equal to 6 89   9467  And now you know why 0 0   8818 is used  it is bitwise different from 0 0   This 15 digits barrier is hard-coded and can only be changed by recompiling the CLI  by using Mono or by calling Microsoft and convincing them to add an option to print full  precision   it is not really precision  but by the lack of a better word   It s probably easier to just calculate the 52 bits precision yourself or use the library mentioned earlier   EDIT  if you like to experiment yourself with IEE-754 floating points  consider this online tool  which shows you all relevant parts of a floating point

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The answer is yes  double printing is broken in  NET  they are printing trailing garbage digits   You can read how to implement it correctly here   I have had to do the same for IronScheme    gt     10 0 0 69  6 8999999999999995  gt  6 89999999999999946709 6 8999999999999995  gt   - 6 9    10 0 0 69   8 881784197001252e-16  gt  6 9 6 9  gt   - 6 9 8 881784197001252e-16  6 8999999999999995   Note  Both C and C  has correct value  just broken printing   Update  I am still looking for the mailing list conversation I had that lead up to this discovery

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The answer to this is simple and can be found on MSDN     Remember that a floating-point number can only approximate a decimal number  and that the precision of a floating-point number determines how accurately that number approximates a decimal number  By default  a Double value contains 15 decimal digits of precision  although a maximum of 17 digits is maintained internally    In your example  the value of i is 6 89999999999999946709  which has the number 9 for all positions between the 3rd and the 16th digit  remember to count the integer part in the digits   When converting to string  the framework rounds the number to the 15th digit   i       6 89999999999999 946709 digit             111111 111122         1 23456789012345 678901

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Digits after decimal point    just two decimal places String Format   0 0 00    123 4567            123 46  String Format   0 0 00    123 4               123 40  String Format   0 0 00    123 0               123 00      max  two decimal places String Format   0 0       123 4567            123 46  String Format   0 0       123 4               123 4  String Format   0 0       123 0               123     at least two digits before decimal point String Format   0 00 0    123 4567            123 5  String Format   0 00 0    23 4567             23 5  String Format   0 00 0    3 4567              03 5  String Format   0 00 0    -3 4567             -03 5   Thousands separator String Format   0 0 0 0    12345 67           12 345 7  String Format   0 0 0    12345 67             12 346   Zero Following code shows how can be formatted a zero  of double type    String Format   0 0 0    0 0                  0 0  String Format   0 0      0 0                  0  String Format   0   0    0 0                   0  String Format   0        0 0                     Align numbers with spaces String Format   0 10 0 0    123 4567               123 5  String Format   0 -10 0 0    123 4567         123 5       String Format   0 10 0 0    -123 4567             -123 5  String Format   0 -10 0 0    -123 4567        -123 5       Custom formatting for negative numbers and zero String Format   0 0 00 minus 0 00 zero    123 4567         123 46  String Format   0 0 00 minus 0 00 zero    -123 4567        minus 123 46  String Format   0 0 00 minus 0 00 zero    0 0              zero   Some funny examples String Format   0 my number is 0 0    12 3         my number is 12 3  String Format   0 0aaa bbb0    12 3

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The problem is that  NET will always round a double to 15 significant decimal digits before applying your formatting  regardless of the precision requested by your format and regardless of the exact decimal value of the binary number   I d guess that the Visual Studio debugger has its own format display routines that directly access the internal binary number  hence the discrepancies between your C  code  your C code and the debugger   There s nothing built-in that will allow you to access the exact decimal value of a double  or to enable you to format a double to a specific number of decimal places  but you could do this yourself by picking apart the internal binary number and rebuilding it as a string representation of the decimal value   Alternatively  you could use Jon Skeet s DoubleConverter class  linked to from his  Binary floating point and  NET  article   This has a ToExactString method which returns the exact decimal value of a double  You could easily modify this to enable rounding of the output to a specific precision   double i   10   0 69  Console WriteLine DoubleConverter ToExactString i    Console WriteLine DoubleConverter ToExactString 6 9 - i    Console WriteLine DoubleConverter ToExactString 6 9        6 89999999999999946709294817992486059665679931640625    0 00000000000000088817841970012523233890533447265625    6 9000000000000003552713678800500929355621337890625

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i tried to reproduce your findings  but when I watched  i  in the debugger it showed up as  6 8999999999999995  not as  6 89999999999999946709  as you wrote in the question  Can you provide steps to reproduce what you saw   To see what the debugger shows you  you can use a DoubleConverter as in the following line of code   Console WriteLine TypeDescriptor GetConverter i  ConvertTo i  typeof string       Hope this helps   Edit  I guess I m more tired than I thought  of course this is the same as formatting to the roundtrip value  as mentioned before

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Take a look at this MSDN reference  In the notes it states that the numbers are rounded to the number of decimal places requested   If instead you use   0 R   it will produce what s referred to as a  round-trip  value  take a look at this MSDN reference for more info  here s my code and the output   double d   10   0 69  Console WriteLine     0 R    d   Console WriteLine     0 F20    6 9 - d   Console WriteLine     0 F20    6 9     output    6 8999999999999995   0 00000000000000088818   6 90000000000000000000

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Another method  starting with the method  double i    10   0 69   Console Write ToStringFull i             Output 6 89999999999999946709294817 Console Write ToStringFull -6 9          Output -6 90000000000000035527136788 Console Write ToStringFull i - 6 9       Output -0 00000000000000088817841970012523233890533  A Drop-In Function    public static string ToStringFull double value        if  value    0 0  return  quot 0 0 quot       if  double IsNaN value   return  quot NaN quot       if  double IsNegativeInfinity value   return  quot -Inf quot       if  double IsPositiveInfinity value   return  quot  Inf quot        long bits   BitConverter DoubleToInt64Bits value       BigInteger mantissa    bits  amp  0xfffffffffffffL    0x10000000000000L      int exp    int   bits  gt  gt  52   amp  0x7ffL  - 1023      string sign    value  lt  0     quot - quot     quot  quot        if  54  gt  exp                double offset    exp   3 321928094887362358        or  Math Log10 Math Abs value           BigInteger temp   mantissa   BigInteger Pow 10  26 -  int offset   gt  gt   52 - exp           string numberText   temp ToString            int digitsNeeded    int   numberText 0  -  5     10 0 - offset           if  exp  lt  0              return sign    quot 0  quot    new string  0   digitsNeeded    numberText          else             return sign   numberText Insert 1 - digitsNeeded   quot   quot              return sign    mantissa  gt  gt   52 - exp   ToString       How it works To solve this problem I used the BigInteger tools  Large values are simple as they just require left shifting the mantissa by the exponent  For small values we cannot just directly right shift as that would lose the precision bits   We must first give it some extra size by multiplying it by a 10 n and then do the right shifts  After that  we move over the decimal n places to the left  More text code here

[c#] Formatting doubles for output in C#

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