Why are some numbers printed with decimal places?

Question

I'm trying to solve a question that says:

You must make a program in C that shows a sequence as in the example below:

I=0 J=1
I=0 J=2
I=0 J=3
I=0.2 J=1.2
I=0.2 J=2.2
I=0.2 J=3.2
.....
I=2 J=?
I=2 J=?
I=2 J=?

I tried to solve this question using the "for" structure, but at the program output, the last three integers appear with decimal places, and the question requires that integers appear without decimal places:

I=0 J=1
I=0 J=2
I=0 J=3
I=0.2 J=1.2
I=0.2 J=2.2
I=0.2 J=3.2
.....
I=1 J=2
I=1 J=3
I=1 J=4
I=1.2 J=2.2
I=1.2 J=3.2
I=1.2 J=4.2
.....
I=2.0 J=3.0
I=2.0 J=4.0
I=2.0 J=5.0

Why does it happen? Here is my code:

int main() {

  int II;
  float I, J, X, FI;
  X = 1;

  for(I = 0; I <= 2.2; I = I + 0.2){

      for(J = X; J <= X + 2; J = J + 1){

          II = (int) I;     //II = The Integer part of I
          FI = I - II;      //FI = The Fractionary part of I

          if(FI == 0)
              printf("I=%.0f J=%.0f\n", I, J);

              //If the fractionary part is 0, then
              //the number must be printed without
              //decimal places.              

          else
              printf("I=%.1f J=%.1f\n", I, J);

              //If the fractionary part is greater than 0,
              //then the number must be printed with just
              //one decimal place.

      }

      X += 0.2;
  }

  return 0;
}

Answer 1

Floating point numbers aren't always exact, you may need to do something like:

if ((FI > -0.1) || (FI < 0.1))
   printf("I=%.0f J=%.0f\n", I, J);
else
   ...

OR

if (fabs(FI) < 0.1)
   printf("I=%.0f J=%.0f\n", I, J);
else
   ...

It is common practice not to compare floats/doubles for exactness, but within some small difference (epsilon) that depends on the particular application.

As an example from the following C# code:

float f = 1;

for (int i = 0; i < 10; i++)
{
  Console.WriteLine(f.ToString("G9"));
  f += 0.2F;
}

This is the output; you can the series in not exact when stored in a float:

1
1.20000005
1.4000001
1.60000014
1.80000019
2.00000024
2.20000029
2.40000033
2.60000038
2.80000043

Answer 2

Why are some numbers printed with decimal places?

Because the algorithm employed assumes exact decimal floating-point math, yet uses the common binary floating point double.

Instead values near OP's goal were computed.

---

Limit issues caused by trying got do decimal math with binary coded floating-pint numbers.

Since code need to iterate 11 times, with I = [0, 0.2, 0.4, ... 2], use an integer counter for the loop and construct I from it.

// for (I = 0; I <= 2.2; I = I + 0.2) {
for (int i = 0; i <= 10; i++) {
  I = i/5.0;

With only this change the output is

I=0 J=1
I=0 J=2
I=0 J=3
I=0.2 J=1.2
I=0.2 J=2.2
I=0.2 J=3.2
...
I=0.8 J=1.8
I=0.8 J=2.8
I=0.8 J=3.8
I=1 J=2
I=1 J=3
I=1 J=4
I=1.2 J=2.2
I=1.2 J=3.2
I=1.2 J=4.2
...
I=1.8 J=2.8
I=1.8 J=3.8
I=1.8 J=4.8
I=2 J=3
I=2 J=4
I=2 J=5

---

To gain insight into what went wrong, it is useful to see the FP values to much greater precision. E. g.:

int main(void) {
  int II;
  float I, J, X, FI;
  X = 1;
  printf("0.2 (%.20f)\n", 0.2);
  printf("2.2 (%.20f)\n", 2.2);
  for(I = 0; I <= 2.2; I = I + 0.2){
      for(J = X; J <= X + 2; J = J + 1){
          II = (int) I;     //II = The Integer part of I
          FI = I - II;      //FI = The Fractionary part of I
          if(FI == 0)
              printf("I=%.0f J=%.0f (%.20f %.20f)\n", I, J, I,J);
          else
              printf("I=%.1f J=%.1f (%.20f %.20f)\n", I, J, I, J);
      }
      X += 0.2;
  }
}

Output

0.2 (0.20000000000000001110)
2.2 (2.20000000000000017764)
I=0 J=1 (0.00000000000000000000 1.00000000000000000000)
I=0 J=2 (0.00000000000000000000 2.00000000000000000000)
I=0 J=3 (0.00000000000000000000 3.00000000000000000000)
I=0.2 J=1.2 (0.20000000298023223877 1.20000004768371582031)
I=0.2 J=2.2 (0.20000000298023223877 2.20000004768371582031)
I=0.2 J=3.2 (0.20000000298023223877 3.20000004768371582031)
I=0.4 J=1.4 (0.40000000596046447754 1.40000009536743164062)
I=0.4 J=2.4 (0.40000000596046447754 2.40000009536743164062)
I=0.4 J=3.4 (0.40000000596046447754 3.40000009536743164062)
I=0.6 J=1.6 (0.60000002384185791016 1.60000014305114746094)
I=0.6 J=2.6 (0.60000002384185791016 2.60000014305114746094)
I=0.6 J=3.6 (0.60000002384185791016 3.60000014305114746094)
I=0.8 J=1.8 (0.80000001192092895508 1.80000019073486328125)
I=0.8 J=2.8 (0.80000001192092895508 2.80000019073486328125)
I=0.8 J=3.8 (0.80000001192092895508 3.80000019073486328125)
I=1 J=2 (1.00000000000000000000 2.00000023841857910156)
I=1 J=3 (1.00000000000000000000 3.00000023841857910156)
I=1 J=4 (1.00000000000000000000 4.00000000000000000000)
I=1.2 J=2.2 (1.20000004768371582031 2.20000028610229492188)
I=1.2 J=3.2 (1.20000004768371582031 3.20000028610229492188)
I=1.2 J=4.2 (1.20000004768371582031 4.20000028610229492188)
I=1.4 J=2.4 (1.40000009536743164062 2.40000033378601074219)
I=1.4 J=3.4 (1.40000009536743164062 3.40000033378601074219)
I=1.4 J=4.4 (1.40000009536743164062 4.40000057220458984375)
I=1.6 J=2.6 (1.60000014305114746094 2.60000038146972656250)
I=1.6 J=3.6 (1.60000014305114746094 3.60000038146972656250)
I=1.6 J=4.6 (1.60000014305114746094 4.60000038146972656250)
I=1.8 J=2.8 (1.80000019073486328125 2.80000042915344238281)
I=1.8 J=3.8 (1.80000019073486328125 3.80000042915344238281)
I=1.8 J=4.8 (1.80000019073486328125 4.80000019073486328125)
I=2.0 J=3.0 (2.00000023841857910156 3.00000047683715820312)
I=2.0 J=4.0 (2.00000023841857910156 4.00000047683715820312)
I=2.0 J=5.0 (2.00000023841857910156 5.00000047683715820312)

Answer 3

Because float numbers will accumulate some imprecision during the iterations. I have done the debug of your code and in the iteration of the value FI=1, the exact value was 8 trailing zeroes. But in the iteration of FI=2 the exact value had some error, so it wouldn't enter the IF condition.

Answer 4

Use %g. As seen in this post

if(FI == 0) printf("I=%g J=%g\n", I, J); //using %g