Using round() function in c

Question

I'm a bit confused about the round() function in C.

First of all, man says:

SYNOPSIS
   #include <math.h>
   double round(double x);

RETURN VALUE
   These functions return the rounded integer value.
   If x is integral, +0, -0, NaN, or infinite, x itself is returned.

The return value is a double / float or an int?

In second place, I've created a function that first rounds, then casts to int. Latter on my code I use it as a mean to compare doubles

int tointn(double in,int n)
{
    int i = 0;
    i = (int)round(in*pow(10,n));
    return i;
}

This function apparently isn't stable throughout my tests. Is there redundancy here? Well... I'm not looking only for an answer, but a better understanding on the subject.

Answer 1

If you want a long integer to be returned then please use lround:

long int tolongint(double in)
{
    return lround(in));
}

For details please see lround which is available as of the C++ 11 standard.

Answer 2

The wording in the man-page is meant to be read literally, that is in its mathematical sense. The wording "x is integral" means that x is an element of Z, not that x has the data type int.

Casting a double to int can be dangerous because the maximum arbitrary integral value a double can hold is 2^52 (assuming an IEEE 754 conforming binary64 ), the maximum value an int can hold might be smaller (it is mostly 32 bit on 32-bit architectures and also 32-bit on some 64-bit architectures).

If you need only powers of ten you can test it with this little program yourself:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(){
  int i;
  for(i = 0;i < 26;i++){
    printf("%d:\t%.2f\t%d\n",i, pow(10,i), (int)pow(10,i));
  }
  exit(EXIT_SUCCESS);
}

Instead of casting you should use the functions that return a proper integral data type like e.g.: lround(3).