将 float 转换为 int 在 C 中无法正常工作

Question

I tried to solve a task on codewars and happened something strange, in one case casting worked as usual and in the second o got strange behavior. Here is the code:

#include <stdio.h>
#include <string.h>
#include <math.h>
int digital_root(int n) {
  char tmp[30];
  char *ptr;
  while(n > 9){
    sprintf(tmp, "%d", n);
    int len = strlen(tmp);
    float n_tmp = n / (pow(10, len));
    n = 0;

    for(int i = 1; i <=len; i++){
      float t = n_tmp * (pow(10, i));
      printf(" vars = [%f , %d] ", t, (int)t); //this line
      n += (int) t;
      n_tmp -= (int)t/pow(10,i);

    }

  }

  return n;
}

int main()
{
    digital_root(16);

    return 0;
}

And the line printf(" vars = [%f , %d] ", t, (int)t); outputs: vars = [1.600000 , 1] vars = [6.000000 , 5] , what is so strange. Can someone explain this behavior?

Answer 1

Every casting float to int is a rounding, so every time you do it, stop for a second and think: what kind of rounding do I need this time ? Dropping the fractional part? floor()? ceil()? Or rounding positive float towards nearest int? If 5.999999999999 should be 6 , use (int)(t+.5) , not (int)t .

Or you can mimic the printf behavior and round only the smallest float fraction, which can vary (depending on the biggest value you use). (int)(t+.000001) will round both 1.9 to 1 (as it probably should), and 3.9999999999 to 4 (because it is 4 with some tiny number representation errors).

You should however know the size of a fraction which can be an actual, meaningful part of your data. Everything smaller than this fraction must be rounded. It can be either .00000001 or .5 , that depends on your needs and the algorithm itself.