如何在C中找出双数字后小数点后是否有任何数字

Question

I stumbled on one issue while I was implementing in C the given algorithm:

int getNumberOfAllFactors(int number) {

  int counter = 0;
  double sqrt_num = sqrt(number);
  for (int i = 1; i <= sqrt_num; i++) {
      if ( number % i == 0) {
          counter = counter + 2;
      }
  }
  if (number == sqrt_num * sqrt_num)
      counter--;

  return counter;
}

– the reason for second condition – is to make a correction for perfect squares (ie 36 = 6 * 6), however it does not avoid situations (false positives) like this one:

sqrt(91) = 18.027756377319946
18.027756377319946 * 18.027756377319946 = 91.0

So my questions are: how to avoid it and what is the best way in C language to figure out whether a double number has any digits after decimal point? Should I cast square root values from double to integers?

Answer 1

在您的情况下，您可以像这样测试它：

if (sqrt_num == (int)sqrt_num)

Answer 2

You should probably use the modf() family of functions:

 #include <math.h> double modf(double value, double *iptr); 

The modf functions break the argument value into integral and fractional parts, each of which has the same type and sign as the argument. They store the integral part (in floating-point format) in the object pointed to by iptr .

This is more reliable than trying to use direct conversions to int because an int is typically a 32-bit number and a double can usually store far larger integer values (up to 53 bits worth) so you can run into errors unnecessarily. If you decide you must use a conversion to int and are working with double values, at least use long long for the conversion rather than int .

(The other members of the family are modff() which handles float and modfl() which handles long double .)