std::remquo,std::remquof,std::remquol(3)

NAME
std::remquo,std::remquof,std::remquol - std::remquo,std::remquof,std::remquol

Synopsis
   Defined in header <cmath>
   float remquo ( float x, float y, int* quo );
   double remquo ( double x, double y, int* quo ); (since C++11)
                        (until C++23)
   long double remquo ( long double x, long double y,
   int* quo );
   constexpr /* floating-point-type */
               remquo ( /* floating-point-type */ x, (since C++23)
                        /* floating-point-type */ y,
   int* quo ); (1)
   float remquof( float x, float y, int* quo ); (2) (since C++11)
                        (constexpr since C++23)
   long double remquol( long double x, long double y, (3) (since C++11)
   int* quo ); (constexpr since C++23)
   Additional overloads
   Defined in header <cmath>
   template< class Arithmetic1, class Arithmetic2 >
   /* common-floating-point-type */ (A) (since C++11)
                        (constexpr since C++23)
       remquo( Arithmetic1 x, Arithmetic2 y, int* quo
   );
   1-3) Computes the floating-point remainder of the division operation x / y as the
   std::remainder() function does. Additionally, the sign and at least the three of the
   last bits of x / y will be stored in quo, sufficient to determine the octant of the
   result within a period.
   The library provides overloads of std::remquo for all cv-unqualified floating-point
   types as the type of the parameters x and y.
   (since C++23)
   A) Additional overloads are provided for all other combinations of arithmetic types.

Parameters
   x, y - floating-point or integer values
   quo - pointer to int to store the sign and some bits of x / y

Return value
   If successful, returns the floating-point remainder of the division x / y as defined
   in std::remainder, and stores, in *quo, the sign and at least three of the least
   significant bits of x / y (formally, stores a value whose sign is the sign of x / y
   and whose magnitude is congruent modulo 2n
   to the magnitude of the integral quotient of x / y, where n is an
   implementation-defined integer greater than or equal to 3).
   If y is zero, the value stored in *quo is unspecified.
   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 is returned if
   subnormals are supported.
   If y is zero, but the domain error does not occur, zero 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),
     * The current rounding mode has no effect.
     * FE_INEXACT is never raised.
     * 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 either x or y is NaN, NaN is returned.

Notes
   POSIX requires that a domain error occurs if x is infinite or y is zero.
   This function is useful when implementing periodic functions with the period exactly
   representable as a floating-point value: when calculating sin(πx) for a very large
   x, calling std::sin directly may result in a large error, but if the function
   argument is first reduced with std::remquo, the low-order bits of the quotient may
   be used to determine the sign and the octant of the result within the period, while
   the remainder may be used to calculate the value with high precision.
   On some platforms this operation is supported by hardware (and, for example, on
   Intel CPUs, FPREM1 leaves exactly 3 bits of precision in the quotient when
   complete).
   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::remquo(num1, num2,
       quo) has the same effect as std::remquo(static_cast<long
       double>(num1),
                   static_cast<long double>(num2), quo).
     * Otherwise, if num1 and/or num2 has type double or an integer type,
       then std::remquo(num1, num2, quo) has the same effect as (until C++23)
       std::remquo(static_cast<double>(num1),
                   static_cast<double>(num2), quo).
     * Otherwise, if num1 or num2 has type float, then std::remquo(num1,
       num2, quo) has the same effect as
       std::remquo(static_cast<float>(num1),
                   static_cast<float>(num2), quo).
   If num1 and num2 have arithmetic types, then std::remquo(num1, num2,
   quo) has the same effect as std::remquo(static_cast</*
   common-floating-point-type */>(num1),
               static_cast</* common-floating-point-type */>(num2), quo),
   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, (since C++23)
   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
// Run this code
 #include <cfenv>
 #include <cmath>
 #include <iostream>
 #ifndef __GNUC__
 #pragma STDC FENV_ACCESS ON
 #endif
 const double pi = std::acos(-1); // or std::numbers::pi since C++20
 double cos_pi_x_naive(double x)
 {
     return std::cos(pi * x);
 }
 // the period is 2, values are (0;0.5) positive, (0.5;1.5) negative, (1.5,2) positive
 double cos_pi_x_smart(double x)
 {
     int quadrant;
     double rem = std::remquo(x, 1, &quadrant);
     quadrant = static_cast<unsigned>(quadrant) % 2; // The period is 2.
     return quadrant == 0 ? std::cos(pi * rem)
                        : -std::cos(pi * rem);
 }
 int main()
 {
     std::cout << std::showpos
               << "naive:\n"
               << " cos(pi * 0.25) = " << cos_pi_x_naive(0.25) << '\n'
               << " cos(pi * 1.25) = " << cos_pi_x_naive(1.25) << '\n'
               << " cos(pi * 2.25) = " << cos_pi_x_naive(2.25) << '\n'
               << "smart:\n"
               << " cos(pi * 0.25) = " << cos_pi_x_smart(0.25) << '\n'
               << " cos(pi * 1.25) = " << cos_pi_x_smart(1.25) << '\n'
               << " cos(pi * 2.25) = " << cos_pi_x_smart(2.25) << '\n'
               << "naive:\n"
               << " cos(pi * 1000000000000.25) = "
               << cos_pi_x_naive(1000000000000.25) << '\n'
               << " cos(pi * 1000000000001.25) = "
               << cos_pi_x_naive(1000000000001.25) << '\n'
               << "smart:\n"
               << " cos(pi * 1000000000000.25) = "
               << cos_pi_x_smart(1000000000000.25) << '\n'
               << " cos(pi * 1000000000001.25) = "
               << cos_pi_x_smart(1000000000001.25) << '\n';
     // error handling
     std::feclearexcept(FE_ALL_EXCEPT);
     int quo;
     std::cout << "remquo(+Inf, 1) = " << std::remquo(INFINITY, 1, &quo) << '\n';
     if (fetestexcept(FE_INVALID))
         std::cout << " FE_INVALID raised\n";
 }

Possible output:
 naive:
   cos(pi * 0.25) = +0.707107
   cos(pi * 1.25) = -0.707107
   cos(pi * 2.25) = +0.707107
 smart:
   cos(pi * 0.25) = +0.707107
   cos(pi * 1.25) = -0.707107
   cos(pi * 2.25) = +0.707107
 naive:
   cos(pi * 1000000000000.25) = +0.707123
   cos(pi * 1000000000001.25) = -0.707117
 smart:
   cos(pi * 1000000000000.25) = +0.707107
   cos(pi * 1000000000001.25) = -0.707107
 remquo(+Inf, 1) = -nan
   FE_INVALID raised

See also
   div(int)
   ldiv computes quotient and remainder of integer division
   lldiv (function)
   (C++11)
   fmod
   fmodf remainder of the floating point division operation
   fmodl (function)
   (C++11)
   (C++11)
   remainder
   remainderf
   remainderl signed remainder of the division operation
   (C++11) (function)
   (C++11)
   (C++11)
   C documentation for
   remquo