使用C语言计算两个浮点数或双精度数的模数

remainder(x, y) = x – rquote * y

#include 
#include 
main() {
   double x = 14.5, y = 4.1;
   double res = remainder(x, y);
   printf("Remainder of %lf/%lf is: %lf
",x,y, res);
   x = -34.50;
   y = 4.0;
   res = remainder(x, y);
   printf("Remainder of %lf/%lf is: %lf
",x,y, res);
   x = 65.23;
   y = 0;
   res = remainder(x, y);
   printf("Remainder of %lf/%lf is: %lf
",x,y, res);
}

Remainder of 14.500000/4.100000 is: -1.900000
Remainder of -34.500000/4.000000 is: 1.500000
Remainder of 65.230000/0.000000 is: -1.#IND00