Argarak - 1 year ago 79
C Question

# C - Comparing floats with if statements

I am currently writing a greedy algorithm, but I have stumbled upon a problem when comparing floats.

I would use code like this:

``````float f = 0;

if (f == 0) {
// code
}
``````

I have tested this on a seperate program and it worked fine, but not on the program I am working on.

Here is an extract from my own program.

``````float chf2 = fmod(chf, 0.1);
float ch3 = chf - chf2;

if (chf2 == 0) {

/* Divide user's number by 0.1 */

float ch3 = chf / 0.1;

/* Round the number */

int ch4 = round(ch3);

/* Print the amount of coins and end */

printf("%d\n", ch4 + coin2);
return 0;
}
``````

Oddly, this seems to work with a previous if statement that checks when a fmod of 0.25 from the user's input.

Is there a better way of checking if a float is equal to another float?

The `fmod` function always returns an exact result---when you write `c = fmod(a, b)`, `c` is a number such that `c + k*b` (evaluated in infinite precision) exactly equals `a` for some integer `k`. So your code is actually sound---`if (c == 0)` will trigger exactly when `a` is exactly a multiple of `b`.
You took `b` to be the `double` `0.1`, which is actually the fraction `3602879701896397/36028797018963968`, not the fraction `1/10`. So if I compute `fmod(1, 0.1)`, I should expect to get the fraction `(36028797018963968 % 3602879701896397) / 36028797018963968`, which is `3602879701896395 36028797018963968`. That's very slightly smaller than the `double` called `0.1`.
This also explains why your code meets your expectations when you use `0.25` instead of `0.1`; `0.25` gives you 1/4 exactly.