NAME¶

std::fmod,std::fmodf,std::fmodl - std::fmod,std::fmodf,std::fmodl

Synopsis¶

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

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

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

fmod ( /* floating-point-type */ x, (since C++23)
(1)
/* floating-point-type */ y );
float fmodf( float x, float y ); (2) (since C++11)
(constexpr since C++23)
long double fmodl( 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 fmod ( Integer x, Integer y );

1-3) Computes the floating-point remainder of the division operation x / y.
The library provides overloads of std::fmod 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 floating-point remainder of the division operation x / y calculated by this
function is exactly the value x - rem * y, where rem is x / y with its fractional
part truncated.

The returned value has the same sign as x and is less than y in magnitude.

Parameters¶

x, y - floating-point or integer values

Return value¶

If successful, returns the 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 (after rounding) 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),

* If x is ±0 and y is not zero, ±0 is returned.
* 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 y is ±∞ and x is finite, x is returned.
* 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 double version of std::fmod behaves as if implemented as follows:

double fmod(double x, double y)
{
#pragma STDC FENV_ACCESS ON
double result = std::remainder(std::fabs(x), y = std::fabs(y));
if (std::signbit(result))
result += y;
return std::copysign(result, x);
}

The expression x - std::trunc(x / y) * y may not equal std::fmod(x, y), when the
rounding of x / y to initialize the argument of std::trunc loses too much precision
(example: x = 30.508474576271183309, y = 6.1016949152542370172).

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::fmod(num1, num2)
has the same effect as std::fmod(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::fmod(num1, num2) has the same effect as (until C++23)
std::fmod(static_cast<double>(num1),
static_cast<double>(num2)).
* Otherwise, if num1 or num2 has type float, then std::fmod(num1,
num2) has the same effect as std::fmod(static_cast<float>(num1),
static_cast<float>(num2)).
If num1 and num2 have arithmetic types, then std::fmod(num1, num2) has
the same effect as std::fmod(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, arguments of (since C++23)
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 << "fmod(+5.1, +3.0) = " << std::fmod(5.1, 3) << '\n'
<< "fmod(-5.1, +3.0) = " << std::fmod(-5.1, 3) << '\n'
<< "fmod(+5.1, -3.0) = " << std::fmod(5.1, -3) << '\n'
<< "fmod(-5.1, -3.0) = " << std::fmod(-5.1, -3) << '\n';

// special values
std::cout << "fmod(+0.0, 1.0) = " << std::fmod(0, 1) << '\n'
<< "fmod(-0.0, 1.0) = " << std::fmod(-0.0, 1) << '\n'
<< "fmod(5.1, Inf) = " << std::fmod(5.1, INFINITY) << '\n';

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

Possible output:¶

fmod(+5.1, +3.0) = 2.1
fmod(-5.1, +3.0) = -2.1
fmod(+5.1, -3.0) = 2.1
fmod(-5.1, -3.0) = -2.1
fmod(+0.0, 1.0) = 0
fmod(-0.0, 1.0) = -0
fmod(5.1, Inf) = 5.1
fmod(+5.1, 0) = -nan
FE_INVALID raised

See also¶

div(int)
ldiv computes quotient and remainder of integer division
lldiv (function)
(C++11)
remainder
remainderf
remainderl signed remainder of the division operation
(C++11) (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
fmod