C - Comparing floats with if statements [duplicate]

问题

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?

回答1:

Your code works correctly. Your expectations might need a little bit of adjusting, however.

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.

You might read this floating-point guide to learn about the representation of floating-point numbers.

来源：https://stackoverflow.com/questions/29084695/c-comparing-floats-with-if-statements

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floating-point

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