is remainder(x,y) really x-((round(x/y)*y)?

Question

Reading the answer of @anton in this link I tried to see if really remainder(x, y) is exactly x-(round(x/y)*y).

Running the code for the value of x=5. and y=2.. I got:

printf("the value of remainder is %f\n",remainder(x, y));
printf("the value of remainder is %f\n",x-(round(x/y)*y));

the value of remainder is 1.000000

the value of remainder is -1.000000

From wikipedia :

Floating point remainder. This is not like a normal modulo operation, it can be negative for two positive numbers. It returns the exact value of x–(round(x/y)·y).

Is the explanation of Anton wrong, or am I missing something ?

Answer 1

There is a slight difference in what remainder does. From the man page:

The remainder() function computes the remainder of dividing x by y. The return value is x-n*y, where n is the value x / y, rounded to the nearest integer. If the absolute value of x-n*y is 0.5, n is chosen to be even.

So in the halfway case the rounding part performed by remainder does not round away from zero, but instead rounds to the nearest even number.