Is there a function to round a float in C or do I need to write my own?

Is there a function to round a float in C or do I need to write my own?

float conver= 45.592346543;

Would like to round actual value to one decimal place. conver = 45.6

Thanks.

From stackoverflow

• #include "math.h" // can't input less than and greater than :)
  
  double round(double x);
  float roundf(float x);
  

  Don't forget to link with -lm. See also ceil(), floor() and trunc().

  Evan Teran : that round to the nearest integer...not what he asked for.

  João da Silva : that's right, I got ahead of myself. I up voted Eduard's answer.

• Sure, you can use roundf(). If you want to round to one decimal, then you could do something like: roundf(10 * x) / 10

  Evan Teran : Nice, everyone else ignored the fact that the asker didn't ask to round to nearest integer. It should be noted that because of imprecision in floating point. When printing you will likely see "45.59999" given the example.

  Evan Teran : A nice more general solution would be: double f(double x, int decimal_points) { int n = pow(10, decimal_points); return roundf(n * x) / n; }

  Tommy : unresolved external symbol _roundf referenced in function _f have included math.h but it doesn't like roundf() I am in VS .NET, is foundf only for linux?

  Eduard - Gabriel Munteanu : Tommy, roundf() is defined in C99, so every compliant compiler should support it. Perhaps you're not linking with the math library?

  Evan Teran : I don't think the newer visual studio's made any effort to support C99.

  Evan Teran : You could just use this though: floor((10 * x) + 0.5) / 10;

• Just to generalize Rob's answer a little, if you're not doing it on output, you can still use the same interface with sprintf().

  I think there is another way to do it, though. You can try ceil() and floor() to round up and down. A nice trick is to add 0.5, so anything over 0.5 rounds up but anything under it rounds down. ceil() and floor() only work on doubles though.

  EDIT: Also, for floats, you can use truncf() to truncate floats. The same +0.5 trick should work to do accurate rounding.

• As Rob mentioned, you probably just want to print the float to 1 decimal place. In this case, you can do something like the following:

  #include <stdio.h>
  #include <stdlib.h>
  
  int main()
  {
    float conver = 45.592346543;
    printf("conver is %0.1f\n",conver);
    return 0;
  }
  

  If you want to actually round the stored value, that's a little more complicated. For one, your one-decimal-place representation will rarely have an exact analog in floating-point. If you just want to get as close as possible, something like this might do the trick:

  #include <stdio.h>
  #include <stdlib.h>
  #include <math.h>
  
  int main()
  {
    float conver = 45.592346543;
    printf("conver is %0.1f\n",conver);
  
    conver = conver*10.0f;
    conver = (conver > (floor(conver)+0.5f)) ? ceil(conver) : floor(conver);
    conver = conver/10.0f;
  
    //If you're using C99 or better, rather than ANSI C/C89/C90, the following will also work.
    //conver = roundf(conver*10.0f)/10.0f;
  
    printf("conver is now %f\n",conver);
    return 0;
  }
  

  I doubt this second example is what you're looking for, but I included it for completeness. If you do require representing your numbers in this way internally, and not just on output, consider using a fixed-point representation instead.

  Tommy : Rounds down seems to work OK but for example rounding 45.569346543; is 45.599998....or 45.5 with *1.0f....I'm closer thought, need to read floor and ceil again. thanks guys.

  Tommy : It is coming from floating point inaccuracy, I changed to double, works great. Thanks.