Round in C with N digits after decimal point

Question

Possible Duplicate:
Rounding Number to 2 Decimal Places in C

I have not found a function with a signature double round(double d, int digits) like here in c. When i try to build i get a error:

error: too many arguments to function 'round'

How can I round in C with N digits after the decimal point?

Solution 1

Using recursion (which is going to be slow for some values of digits)

#include <math.h>
double my_round(double x, unsigned int digits) {
  if (digits > 0) {
    return my_round(x*10.0, digits-1)/10.0;
  }
  else {
    return round(x);
  }
}

A method likely to be somewhat faster, but which relies on a single call to the slow pow function:

#include <math.h>

double my_round(double x, unsigned int digits) {
    double fac = pow(10, digits);
    return round(x*fac)/fac;
}

An even faster method is to precompute a lookup table with the likely powers and use that instead of pow.

#include <math.h>

double fac[];  // population of this is left as an exercise for the reader

double my_round(double x, unsigned int digits) {
    return round(x*fac[digits])/fac[digits];
}

Solution 2

Whilst "answerd" gives a decent answer, here's one that works for arbitrarily large numbers:

double round1(double num, int N) {
      ASSERT(N > 0);
      double p10 = pow(10,N);
      return round(num* p10) / p10;
}

Of course, as stated, floating point numbers don't have a set number of decimal digits, and this is NOT guaranteed to PRINT as 3.70000 if you call printf("%8.5f", round1(3.7519, 1)); for example.