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

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)

S) The SIMD overload performs an element-wise std::fmod on v_xand v_y.

(See math-common-simd-t for its definition.)

(since C++26)

A) Additional overloads are provided for all integer types, which are treated as double.

(since C++11)

The floating-point remainder of the division operation x / y calculated by this function is exactly the value x - iquot * y, where iquot is x / y with its fractional part truncated.

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

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)).

(until C++23)

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.

(since C++23)

[edit] 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