NAME¶

std::remainder,std::remainderf,std::remainderl - std::remainder,std::remainderf,std::remainderl

Synopsis¶

Defined in header <cmath>
float remainder ( float x, float y );

double remainder ( double x, double y ); (until C++23)

long double remainder ( long double x, long double y
);
constexpr /* floating-point-type */

remainder ( /* floating-point-type */ x, (since C++23)
(1)
/* floating-point-type */ y
);
float remainderf( float x, float y ); (2) (since C++11)
(constexpr since C++23)
long double remainderl( long double x, long double y (3) (since C++11)
); (constexpr since C++23)
Additional overloads (since C++11)
Defined in header <cmath>
template< class Integer > (A) (constexpr since C++23)
double remainder ( Integer x, Integer y );

1-3) Computes the IEEE remainder of the floating point division operation x / y.
The library provides overloads of std::remainder for all cv-unqualified
floating-point types as the type of the parameters.
(since C++23)

A) Additional overloads are provided for all integer types, which are (since C++11)
treated as double.

The IEEE floating-point remainder of the division operation x / y calculated by this
function is exactly the value x - quo * y, where the value quo is the integral value
nearest the exact value x / y. When |quo - x / y| = ½, the value quo is chosen to be
even.

In contrast to std::fmod, the returned value is not guaranteed to have the same sign
as x.

If the returned value is zero, it will have the same sign as x.

Parameters¶

x, y - floating-point or integer values

Return value¶

If successful, returns the IEEE floating-point remainder of the division x / y as
defined above.

If a domain error occurs, an implementation-defined value is returned (NaN where
supported).

If a range error occurs due to underflow, the correct result is returned.

If y is zero, but the domain error does not occur, zero is returned.

Error handling¶

Errors are reported as specified in math_errhandling.

Domain error may occur if y is zero.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

* The current rounding mode has no effect.
* FE_INEXACT is never raised, the result is always exact.
* If x is ±∞ and y is not NaN, NaN is returned and FE_INVALID is raised.
* If y is ±0 and x is not NaN, NaN is returned and FE_INVALID is raised.
* If either argument is NaN, NaN is returned.

Notes¶

POSIX requires that a domain error occurs if x is infinite or y is zero.

std::fmod, but not std::remainder is useful for doing silent wrapping of
floating-point types to unsigned integer types: (0.0 <= (y = std::fmod(std::rint(x),
65536.0)) ? y : 65536.0 + y) is in the range [-0.0, 65535.0], which corresponds to
unsigned short, but std::remainder(std::rint(x), 65536.0) is in the range
[-32767.0, +32768.0], which is outside of the range of signed short.

The additional overloads are not required to be provided exactly as (A). They only
need to be sufficient to ensure that for their first argument num1 and second
argument num2:

* If num1 or num2 has type long double, then std::remainder(num1,
num2) has the same effect as std::remainder(static_cast<long
double>(num1),
static_cast<long double>(num2)).
* Otherwise, if num1 and/or num2 has type double or an integer type,
then std::remainder(num1, num2) has the same effect as (until C++23)
std::remainder(static_cast<double>(num1),
static_cast<double>(num2)).
* Otherwise, if num1 or num2 has type float, then
std::remainder(num1, num2) has the same effect as
std::remainder(static_cast<float>(num1),
static_cast<float>(num2)).
If num1 and num2 have arithmetic types, then std::remainder(num1,
num2) has the same effect as std::remainder(static_cast</*
common-floating-point-type */>(num1),
static_cast</* common-floating-point-type */>(num2)),
where /* common-floating-point-type */ is the floating-point type with
the greatest floating-point conversion rank and greatest
floating-point conversion subrank between the types of num1 and num2, (since C++23)
arguments of integer type are considered to have the same
floating-point conversion rank as double.

If no such floating-point type with the greatest rank and subrank
exists, then overload resolution does not result in a usable candidate
from the overloads provided.

Example¶

// Run this code

#include <cfenv>
#include <cmath>
#include <iostream>
// #pragma STDC FENV_ACCESS ON

int main()
{
std::cout << "remainder(+5.1, +3.0) = " << std::remainder(5.1, 3) << '\n'
<< "remainder(-5.1, +3.0) = " << std::remainder(-5.1, 3) << '\n'
<< "remainder(+5.1, -3.0) = " << std::remainder(5.1, -3) << '\n'
<< "remainder(-5.1, -3.0) = " << std::remainder(-5.1, -3) << '\n';

// special values
std::cout << "remainder(-0.0, 1.0) = " << std::remainder(-0.0, 1) << '\n'
<< "remainder(5.1, Inf) = " << std::remainder(5.1, INFINITY) << '\n';

// error handling
std::feclearexcept(FE_ALL_EXCEPT);
std::cout << "remainder(+5.1, 0) = " << std::remainder(5.1, 0) << '\n';
if (fetestexcept(FE_INVALID))
std::cout << " FE_INVALID raised\n";
}

Possible output:¶

remainder(+5.1, +3.0) = -0.9
remainder(-5.1, +3.0) = 0.9
remainder(+5.1, -3.0) = -0.9
remainder(-5.1, -3.0) = 0.9
remainder(-0.0, 1.0) = -0
remainder(5.1, Inf) = 5.1
remainder(+5.1, 0) = -nan
FE_INVALID raised

See also¶

div(int)
ldiv computes quotient and remainder of integer division
lldiv (function)
(C++11)
fmod
fmodf remainder of the floating point division operation
fmodl (function)
(C++11)
(C++11)
remquo
remquof
remquol signed remainder as well as the three last bits of the division operation
(C++11) (function)
(C++11)
(C++11)
C documentation for
remainder