Fmod does not work as expected(c programming)

Question

(the problem is caused by me and has been solved ,greetings from the newbie)I apologize to everyone, my function type was integer i just realized it, I opened it because I worked for hours, I have to delete it). I am using gcc 9.3.0 version and standard settings. In the code in the example, I am getting 7 while waiting to get output 7.14. What could be the reason for this?(im using gcc main.c -lm -o main when compiling,what I'm trying to do is print as an integer if the double number is equal to integer, otherwise print it as a double,sorry for the adjustments, this is the last and proper version)

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

int try(double a);

int main() {
  double b = 7.14;
  double x;
  double temp;

  x = try(b);

  if (fmod(x, 1) == 0.0) {  // print 7 in this function
    temp = x;
    printf("%d", (int)temp);
  } 
   else if (fmod(x, 1) != 0.0) {
    printf("%f", x);
  }

  return 0;
}

int try(double a) {
  if (fmod(a, 1) != 0.0) {
    printf("%lf", a);  // print 7.14 in this function
  } else if (fmod(a, 1) == 0.0) {
    printf("%d", (int)a);
  }

  return a + 0;
}

Also, it's a bad idea to compare a double with ==. Instead use <= delta where delta is some small value. — Allan Wind, Mar 24 '21 at 01:52
I use the double value returned from the other function instead of x, when I print it in the other function, it prints it as 7.14 but when i do this in main function it print 7.0.I immediately edit the question to avoid confusion — user15465584, Mar 24 '21 at 01:57
I get 7.140000 with gcc 9.3.0. https://wandbox.org/permlink/xipjKx9PUx6fgox0 Same after the edit: https://wandbox.org/permlink/UMHioIMbBkvawokq — Retired Ninja, Mar 24 '21 at 02:00
I believe the printf format specifier for `double` should be just `%f` and not `%lf`. (It's different from scanf.) Does that change anything? — Nate Eldredge, Mar 24 '21 at 02:02
Older compilers have trouble printing with `"%lf"`. Try `"%f"`. — chux - Reinstate Monica, Mar 24 '21 at 02:05
user15465584, " I am getting 7 " --> Did you mean `"7.000000"`, `"7"` or something else? — chux - Reinstate Monica, Mar 24 '21 at 02:06
I couldn't edit the final version right away due to internet problem. I'm sorry. No, it only prints 7 . — user15465584, Mar 24 '21 at 02:09
Hmmm `try` is a reserved word in other languages, perhaps another word? Or the code posted is not the true code. Review the cut/paste and maybe improve format.. — chux - Reinstate Monica, Mar 24 '21 at 02:09
Please don't vandalize your posts. I'm reverting your last change. — ChrisGPT was on strike, Mar 24 '21 at 02:49

score 0 · Answer 1 · answered Mar 24 '21 at 01:51

0

fmod(7.14, 1) = 0.14, thus it hit the first if and print 7.14.

Compute remainder of division

Returns the floating-point remainder of numer/denom (rounded towards zero):

fmod = numer - tquot * denom

Where tquot is the truncated (i.e., rounded towards zero) result of: numer/denom.

http://www.cplusplus.com/reference/cmath/fmod/

answered Mar 24 '21 at 01:51

delta

3,778
15
22

OP says they're getting 7, not 7.14, though. – Shawn Mar 24 '21 at 01:53

score 0 · Accepted Answer · answered Mar 24 '21 at 02:14

With gcc version 10.2.0 (GCC)

Got the result 7.140000

Some code-format and debug output.

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

double func_try(double a) { return a + 0; }

int main() {
  double b = 7.14;
  double x;
  x = func_try(b);

  printf("%lf\n", fmod(x, 1));

  if (fmod(x, 1) != 0.0) {
    printf("%lf", x);
  } else if (fmod(x, 1) == 0.0) {
    printf("%d", (int)x);
  }

  return 0;
}

$gcc main.c -lm -o main

$./main
0.140000
7.140000

Fmod does not work as expected(c programming)

2 Answers2