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 ?

The formula assumes that x and y are floating point, not integers. — Raymond Chen, Aug 03 '16 at 15:39
@RaymondChen -_- if they were integers I will get the correct answer ! No comment ! — rondino, Aug 03 '16 at 15:40
IEEE uses banker's rounding. They think round(2.5) should be 2 and not 3. — Hans Passant, Aug 03 '16 at 15:46
@HansPassant is `nearbyint()` function using banker's rounding ? — rondino, Aug 03 '16 at 15:52
It uses the rounding mode set by std::fesetround(). No bankers in the C++ standardization committee. — Hans Passant, Aug 03 '16 at 16:00
Sorry, what's the meaning of `round()` in the wikipedia context? can you edit your question and add a definition of `round()` function? — Luis Colorado, Aug 05 '16 at 07:29
check the hans Passant comment which is above. It is banker's rounding — rondino, Aug 07 '16 at 15:34

dbush · Accepted Answer · 2016-08-03T16:04:19.773

2

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.

edited Aug 03 '16 at 16:04

answered Aug 03 '16 at 15:44

dbush

205,898
23
218
273

so I should use `nearbyint()` function instead ? – rondino Aug 03 '16 at 15:46
1

Good answer (+1). Might want to add in a link/reference to [banker's rounding](http://c2.com/cgi/wiki?BankersRounding) so OP can learn more about different rounding algorithms and their error bounds. :) – Cloud Aug 03 '16 at 15:46
I will check your answer if you provide a link for that ! – rondino Aug 03 '16 at 16:00
@rondino Added the link. Also changed quoted test as the wording differs from my local man page (the meaning is the same, though). – dbush Aug 03 '16 at 16:06

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

1 Answers1