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

Since C++23

Header: <cmath>

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)

# Declarations

float fmod ( float x, float y );
double fmod ( double x, double y );
long double fmod ( long double x, long double y );

(until C++23)

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

(since C++23)

float fmodf( float x, float y );

(since C++11) (constexpr since C++23)

long double fmodl( long double x, long double y );

(since C++11) (constexpr since C++23)

SIMD overload (since C++26)

template< class V0, class V1 >
constexpr /*math-common-simd-t*/<V0, V1>
fmod ( const V0& v_x, const V1& v_y );

(since C++26)

Additional overloads (since C++11)

template< class Integer >
double fmod ( Integer x, Integer y );

(constexpr since C++23)

# Parameters

# Return value

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

# 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:

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 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.

# Example

#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";
}

# See also