How do I restrict a float value to only two places after the decimal point in C

Question

How can I round a float value  such as 37 777779  to two decimal places  37 78  in C

User · Answer

If you just want to round the number for output purposes, then the "%.2f" format string is indeed the correct answer. However, if you actually want to round the floating point value for further computation, something like the following works:

#include <math.h>

float val = 37.777779;

float rounded_down = floorf(val * 100) / 100;   /* Result: 37.77 */
float nearest = roundf(val * 100) / 100;  /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100;      /* Result: 37.78 */

Notice that there are three different rounding rules you might want to choose: round down (ie, truncate after two decimal places), rounded to nearest, and round up. Usually, you want round to nearest.

As several others have pointed out, due to the quirks of floating point representation, these rounded values may not be exactly the "obvious" decimal values, but they will be very very close.

For much (much!) more information on rounding, and especially on tie-breaking rules for rounding to nearest, see the Wikipedia article on Rounding.

User · Answer

Use float roundf float x     The round functions round their argument to the nearest integer value in floating-point format  rounding halfway cases away from zero  regardless of the current rounding direction    C11dr   7 12 9 5   include  lt math h gt  float y   roundf x   100 0f    100 0f       Depending on your float implementation  numbers that may appear to be half-way are not  as floating-point is typically base-2 oriented    Further  precisely rounding to the nearest 0 01 on all  half-way  cases is most challenging   void r100 const char  s      float x  y    sscanf s    f    amp x     y   round x 100 0  100 0    printf   6s   12e   12e n   s  x  y      int main void      r100  1 115      r100  1 125      r100  1 135      return 0      1 115 1 115000009537e 00 1 120000004768e 00    1 125 1 125000000000e 00 1 129999995232e 00  1 135 1 134999990463e 00 1 139999985695e 00   Although  1 115  is  half-way  between 1 11 and 1 12  when converted to float  the value is 1 115000009537    and is no longer  half-way   but closer to 1 12 and rounds to the closest float of 1 120000004768      1 125  is  half-way  between 1 12 and 1 13  when converted to float  the value is exactly 1 125 and is  half-way   It rounds toward 1 13 due to ties to even rule and rounds to the closest float of 1 129999995232     Although  1 135  is  half-way  between 1 13 and 1 14  when converted to float  the value is 1 134999990463    and is no longer  half-way   but closer to 1 13 and rounds to the closest float of 1 129999995232     If code used   y   roundf x 100 0f  100 0f    Although  1 135  is  half-way  between 1 13 and 1 14  when converted to float  the value is 1 134999990463    and is no longer  half-way   but closer to 1 13 but incorrectly rounds to float of 1 139999985695    due to the more limited precision of float vs  double   This incorrect value may be viewed as correct  depending on coding goals

User · Answer

Using   2f in printf  It only print 2 decimal points   Example   printf    2f   37 777779     Output   37 77

User · Answer

In C    or in C with C-style casts   you could create the function      Function to control   of decimal places to be output for x    double showDecimals const double amp  x  const int amp  numDecimals        int y x      double z x-y      double m pow 10 numDecimals       double q z m      double r round q        return static cast lt double gt  y   1 0 m  r      Then std  cout  lt  lt  showDecimals 37 777779 2   would produce  37 78   Obviously you don t really need to create all 5 variables in that function  but I leave them there so you can see the logic   There are probably simpler solutions  but this works well for me--especially since it allows me to adjust the number of digits after the decimal place as I need

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or you can do it the old-fashioned way without any libraries   float a   37 777779   int b   a     b   37     float c   a - b     c   0 777779    c    100     c   77 777863    int d   c     d   77      a   b   d    float 100     a   37 770000    That of course if you want to remove the extra information from the number

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Also  if you re using C    you can just create a function like this   string prd const double x  const int decDigits        stringstream ss      ss  lt  lt  fixed      ss precision decDigits      set   places after decimal     ss  lt  lt  x      return ss str        You can then output any double myDouble with n places after the decimal point with code such as this   std  cout  lt  lt  prd myDouble n

User · Answer

There isn t a way to round a float to another float because the rounded float may not be representable  a limitation of floating-point numbers    For instance  say you round 37 777779 to 37 78  but the nearest representable number is 37 781   However  you can  round  a float by using a format string function

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this function takes the number and precision and returns the rounded off number  float roundoff float num int precision          int temp  int   num pow 10 precision          int num1 num pow 10 precision 1         temp  10        temp  5        if num1 gt  temp                num1  10        num1  10        num1  10        num num1 pow 10 precision 1         return num      it converts the floating point number into int by left shifting the point and checking for the greater than five condition

User · Answer

Assuming you re talking about round the value for printing  then Andrew Coleson and AraK s answer are correct   printf    2f   37 777779     But note that if you re aiming to round the number to exactly 37 78 for internal use  eg to compare against another value   then this isn t a good idea  due to the way floating point numbers work  you usually don t want to do equality comparisons for floating point  instead use a target value   - a sigma value   Or encode the number as a string with a known precision  and compare that   See the link in Greg Hewgill s answer to a related question  which also covers why you shouldn t use floating point for financial calculations

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How about this   float value   37 777779  float rounded     int  value   100    5    100 0

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double f round double dval  int n        char l fmtp 32   l buf 64       char  p str      sprintf  l fmtp       df   n       if  dval gt  0              sprintf  l buf  l fmtp  dval       else             sprintf  l buf  l fmtp  dval       return   double strtod l buf   amp p str         Here n is the number of decimals  example   double d   100 23456   printf   f   f round d  4      result  100 2346  printf   f   f round d  2      result  100 23

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Always use the printf family of functions for this  Even if you want to get the value as a float  you re best off using snprintf to get the rounded value as a string and then parsing it back with atof    include  lt math h gt   include  lt stdio h gt   include  lt stddef h gt   include  lt stdlib h gt   double dround double val  int dp        int charsNeeded   1   snprintf NULL  0      f   dp  val       char  buffer   malloc charsNeeded       snprintf buffer  charsNeeded      f   dp  val       double result   atof buffer       free buffer       return result      I say this because the approach shown by the currently top-voted answer and several others here -  multiplying by 100  rounding to the nearest integer  and then dividing by 100 again - is flawed in two ways    For some values  it will round in the wrong direction because the multiplication by 100 changes the decimal digit determining the rounding direction from a 4 to a 5 or vice versa  due to the imprecision of floating point numbers For some values  multiplying and then dividing by 100 doesn t round-trip  meaning that even if no rounding takes place the end result will be wrong   To illustrate the first kind of error - the rounding direction sometimes being wrong - try running this program   int main void           This number is EXACTLY representable as a double     double x   0 01499999999999999944488848768742172978818416595458984375       printf  x    50f n   x        double res1   dround x  2       double res2   round 100   x    100       printf  Rounded with snprintf    50f n   res1       printf  Rounded with round  then divided    50f n   res2       You ll see this output   x  0 01499999999999999944488848768742172978818416595459 Rounded with snprintf  0 01000000000000000020816681711721685132943093776703 Rounded with round  then divided  0 02000000000000000041633363423443370265886187553406   Note that the value we started with was less than 0 015  and so the mathematically correct answer when rounding it to 2 decimal places is 0 01  Of course  0 01 is not exactly representable as a double  but we expect our result to be the double nearest to 0 01  Using snprintf gives us that result  but using round 100   x    100 gives us 0 02  which is wrong  Why  Because 100   x gives us exactly 1 5 as the result  Multiplying by 100 thus changes the correct direction to round in   To illustrate the second kind of error - the result sometimes being wrong due to   100 and   100 not truly being inverses of each other - we can do a similar exercise with a very big number   int main void        double x   8631192423766613 0       printf  x    1f n   x        double res1   dround x  2       double res2   round 100   x    100       printf  Rounded with snprintf    1f n   res1       printf  Rounded with round  then divided    1f n   res2       Our number now doesn t even have a fractional part  it s an integer value  just stored with type double  So the result after rounding it should be the same number we started with  right   If you run the program above  you ll see   x  8631192423766613 0 Rounded with snprintf  8631192423766613 0 Rounded with round  then divided  8631192423766612 0   Oops  Our snprintf method returns the right result again  but the multiply-then-round-then-divide approach fails  That s because the mathematically correct value of 8631192423766613 0   100  863119242376661300 0  is not exactly representable as a double  the closest value is 863119242376661248 0  When you divide that back by 100  you get 8631192423766612 0 - a different number to the one you started with   Hopefully that s a sufficient demonstration that using roundf for rounding to a number of decimal places is broken  and that you should use snprintf instead  If that feels like a horrible hack to you  perhaps you ll be reassured by the knowledge that it s basically what CPython does

User · Answer

You can still use   float ceilf float x      don t forget  include  lt math h gt  and link with -lm    example   float valueToRound   37 777779  float roundedValue   ceilf valueToRound   100    100

User · Answer

Let me first attempt to justify my reason for adding yet another answer to this question   In an ideal world  rounding is not really a big deal   However  in real systems  you may need to contend with several issues that can result in rounding that may not be what you expect   For example  you may be performing financial calculations where final results are rounded and displayed to users as 2 decimal places  these same values are stored with fixed precision in a database that may include more than 2 decimal places  for various reasons  there is no optimal number of places to keep   depends on specific situations each system must support  e g  tiny items whose prices are fractions of a penny per unit   and  floating point computations performed on values where the results are plus minus epsilon   I have been confronting these issues and evolving my own strategy over the years   I won t claim that I have faced every scenario or have the best answer  but below is an example of my approach so far that overcomes these issues   Suppose 6 decimal places is regarded as sufficient precision for calculations on floats doubles  an arbitrary decision for the specific application   using the following rounding function method   double Round double x  int p        if  x    0 0            return   floor  fabs x  pow double 10 0  p   0 5   pow double 10 0  p    x fabs x          else           return 0 0            Rounding to 2 decimal places for presentation of a result can be performed as   double val        perform calculations on val String Round Round Round val 8  6  2      For val   6 825  result is 6 83 as expected   For val   6 824999  result is 6 82   Here the assumption is that the calculation resulted in exactly 6 824999 and the 7th decimal place is zero   For val   6 8249999  result is 6 83   The 7th decimal place being 9 in this case causes the Round val 6  function to give the expected result   For this case  there could be any number of trailing 9s   For val   6 824999499999  result is 6 83   Rounding to the 8th decimal place as a first step  i e  Round val 8   takes care of the one nasty case whereby a calculated floating point result calculates to 6 8249995  but is internally represented as 6 824999499999      Finally  the example from the question   val   37 777779 results in 37 78   This approach could be further generalized as   double val        perform calculations on val String Round Round Round val N 2  N  2      where N is precision to be maintained for all intermediate calculations on floats doubles   This works on negative values as well   I do not know if this approach is mathematically correct for all possibilities

User · Answer

printf    2f   37 777779     If you want to write to C-string   char number 24      dummy size  you should take care of the size  sprintf number     2f   37 777779

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I made this macro for rounding float numbers  Add it in your header   being of file   define ROUNDF f  c     float   int   f     c       c      Here is an example   float x   ROUNDF 3 141592  100    x equals 3 14

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Code definition     define roundz x d    floor   x  pow 10 d    5   pow 10 d     Results    a   8 000000 sqrt a    r   2 828427 roundz r 2    2 830000 roundz r 3    2 828000 roundz r 5    2 828430

[c] How do I restrict a float value to only two places after the decimal point in C?

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