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std::fmod,std::fmodf,std::fmodl(3) | C++ Standard Libary | std::fmod,std::fmodf,std::fmodl(3) |

# 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

2024.06.10 | http://cppreference.com |