Fmod

Defined in header <cmath>.

Description

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.

Declarations

C++23

// 1)
constexpr /* floating-point-type */
            fmod ( /* floating-point-type */ x,
                   /* floating-point-type */ y );
// 2)
constexpr float       fmodf( float x, float y );
// 3)
constexpr long double fmodl( long double x, long double y );

Additional Overloads

// 4)
template< class Arithmetic1, class Arithmetic2 >
/* common-floating-point-type */ fmod( Arithmetic1 x, Arithmetic2 y );

C++11

// 1)
float       fmod ( float x, float y );
double      fmod ( double x, double y );
long double fmod ( long double x, long double y );
// 2)
float       fmodf( float x, float y );
// 3)
long double fmodl( long double x, long double y );

Additional Overloads

// 4)
template< class Arithmetic1, class Arithmetic2 >
/* common-floating-point-type */ fmod( Arithmetic1 x, Arithmetic2 y );

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

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
​ 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 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.

Examples

#include <iostream>
#include <cmath>
#include <cfenv>
 
// #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";
}

Result

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