如何在 C 中找到双精度数的余数？ 模仅适用于整数

Question

This is what I've found so far online,

int main(void)
{
    long a = 12345;
    int b = 10;

    int remain = a - (a / b) * b;
    printf("%i\n", remain);
}

First I wonder how the formula works. Maybe i cant do math, but the priority of operations here seems a bit odd. If i run this code the expected answer of 5 is printed. But I dont get how (a / b) * b doesn't cancel out to 'a' leading to a - a = 0.

Now, this only works for int and long, as soon as double are involved it doesn't work anymore. Anyone might tell me why? Is there an alternative to modulo that works for double?

Also I'm not sure if i understand up to what value a long can go, i found online that the upper limit was 2147483647 but when i input bigger numbers such as the one in 'a' the code runs without any issue up to a certain point...

Thanks for your help I'm new to coding and trying to learn!

Answer 1

Given two double finite numbers x and y , with y not equal to zero, fmod(x, y) produces the remainder of x when divided by y . Specifically, it returns x − n y , where n is chosen so that x − n y has the same sign as x and is smaller in magnitude than y . (So, if x is positive, 0 ≤ fmod(x, y) < x , and, if x is negative, x < fmod(x, y) ≤ 0.)

fmod is declared in <math.h> .

A properly implemented fmod returns an exact result; there is no floating-point error, since the specified result is always representable.

The C standard also specifies remquo to return the remainder and some low bits (at least three) of the quotient n and remainder with a variation on the definition of the remainder. It also specifies variants of these functions for float and long double .

Answer 2

Naive implementation. Limited range. Adds additional floating point imprecisions (as it does some arithmetic)

double naivemod(double x)
{
    return x - (long long)x;
}


int main(void)
{
    printf("%.50f\n", naivemod(345345.567567756));
    printf("%.50f\n", naivemod(.0));
    printf("%.50f\n", naivemod(10.5));
    printf("%.50f\n", naivemod(-10.0/3));
    
}