Understanding float variable comparison in if()

Question

Unable to find the reason for the following piece of code:

#include <stdio.h>
int main()
{
    float f = 0.1;
    if (f == 0.1)
      printf("True");
    else
      printf("False");
    return 0;
}

The output is false.

#include <stdio.h>
int main()
{
    float f = 0.1;
    if (f == (float)0.1)
      printf("True");
    else
      printf("False");
    return 0;
}

Now shows the correct output. Whats the reason behind this?

Also what is the reason of this behavior.

#include <stdio.h>
main()
{
    int n = 0, m = 0;
        if (n > 0)
            if (m > 0)
                    printf("True");
        else 
            printf("False");
}

Answer 1

0.1 literal is double. You loose precision here float f = 0.1;

You could say we loose precision during comparison again, so why isn't f == 0.1 true anyway? Because float extends to double, not the opposite. In C smaller type always extends to the bigger one.

Simplified your example we can say that double(float(1.0)) != 1.0

Possible solutions:

• use double instead of float as a type of f.
• use cast as in your 2nd example
• use float literals - replace all 0.1 with 0.1f

Better solution

Floating point variables have a lot of problems with comparisons. They, including this one, can be solved by defining your own comparison function:

bool fp_equal(double a, double b, double eps = FLT_EPSILON) {
  return fabs(a - b) < fabs(eps * a);
}

---

The second part of the question:

Why the answer is false is because else part always corresponds to the innermost if block. So you were confused by formatting, the code is equivalent to:

#include <stdio.h>

int main()
{
    int n = 0, m = 0;
    if (n > 0) {
        if (m > 0) {
            printf("True");
        }
        else {
            printf("False");
        }
    }
}