Convert float to double without losing precision

Question

I have a primitive float and I need as a primitive double  Simply casting the float to double gives me weird extra precision  For example   float temp   14009 35F  System out println Float toString temp       Prints 14009 35 System out println Double toString  double temp       Prints 14009 349609375   However  if instead of casting  I output the float as a string  and parse the string as a double  I get what I want   System out println Double toString Double parseDouble Float toString temp         Prints 14009 35   Is there a better way than to go to String and back

User · Answer

Use a BigDecimal instead of float double  There are a lot of numbers which can t be represented as binary floating point  for example  0 1   So you either must always round the result to a known precision or use BigDecimal   See http   en wikipedia org wiki Floating point for more information

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It s not that you re actually getting extra precision - it s that the float didn t accurately represent the number you were aiming for originally  The double is representing the original float accurately  toString is showing the  extra  data which was already present   For example  and these numbers aren t right  I m just making things up  suppose you had   float f   0 1F  double d   f    Then the value of f might be exactly 0 100000234523  d will have exactly the same value  but when you convert it to a string it will  trust  that it s accurate to a higher precision  so won t round off as early  and you ll see the  extra digits  which were already there  but hidden from you   When you convert to a string and back  you re ending up with a double value which is closer to the string value than the original float was - but that s only good if you really believe that the string value is what you really wanted   Are you sure that float double are the appropriate types to use here instead of BigDecimal  If you re trying to use numbers which have precise decimal values  e g  money   then BigDecimal is a more appropriate type IMO

User · Answer

I found the following solution   public static Double getFloatAsDouble Float fValue        return Double valueOf fValue toString         If you use float and double instead of Float and Double use the following   public static double getFloatAsDouble float value        return Double valueOf Float valueOf value  toString    doubleValue

User · Answer

A simple solution that works well  is to parse the double from the string representation of the float   double val   Double valueOf String valueOf yourFloat      Not super efficient  but it works

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For information this comes under Item 48 - Avoid float and double when exact values are required  of Effective Java 2nd edition by Joshua Bloch  This book is jam packed with good stuff and definitely worth a look

User · Answer

Does this work   float flt   145 664454   Double dbl   0 0  dbl    flt

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Floats  by nature  are imprecise and always have neat rounding  issues    If precision is important then you might consider refactoring your application to use Decimal or BigDecimal     Yes  floats are computationally faster than decimals because of the on processor support   However  do you want fast or accurate

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This is due the contract of Float toString float   which says in part       How many digits must be printed for   the fractional part   hellip    There   must be at least one digit to   represent the fractional part  and   beyond that as many  but only as many    more digits as are needed to uniquely   distinguish the argument value from   adjacent values of type float  That   is  suppose that x is the exact   mathematical value represented by the   decimal representation produced by   this method for a finite nonzero   argument f  Then f must be the float    value nearest to x  or  if two float   values are equally close to x  then f   must be one of them and the least   significant bit of the significand of   f must be 0

User · Answer

I find converting to the binary representation easier to grasp this problem   float f   0 27f  double d2    double  f  double d3   0 27d   System out println Integer toBinaryString Float floatToRawIntBits f     System out println Long toBinaryString Double doubleToRawLongBits d2     System out println Long toBinaryString Double doubleToRawLongBits d3       You can see the float is expanded to the double by adding 0s to the end  but that the double representation of 0 27 is  more accurate   hence the problem      111110100010100011110101110001 11111111010001010001111010111000100000000000000000000000000000 11111111010001010001111010111000010100011110101110000101001000

User · Answer

I ve encountered this issue today and could not use refactor to BigDecimal  because the project is really huge  However I found solution using  Float result   new Float 5623 23  Double doubleResult   new FloatingDecimal result floatValue    doubleValue     And this works   Note that calling result doubleValue   returns 5623 22998046875  But calling doubleResult doubleValue   returns correctly 5623 23  But I am not entirely sure if its a correct solution

[java] Convert float to double without losing precision

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