# std::remquo, std::remquof, std::remquol

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< cpp‎ | numeric‎ | math

Defined in header `<cmath>` |  |
| (1) |
float remquo ( float x, float y, int* quo ); double remquo ( double x, double y, int* quo ); long double remquo ( long double x, long double y, int* quo ); |  | (since C++11) (until C++23)
constexpr /* floating-point-type */ remquo ( /* floating-point-type */ x, /* floating-point-type */ y, int* quo ); |  | (since C++23)
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, int* quo ); | (3) | (since C++11) (constexpr since C++23)
Additional overloads |  |
Defined in header `<cmath>` |  |
template< class Arithmetic1, class Arithmetic2 > /* common-floating-point-type */ remquo( Arithmetic1 x, Arithmetic2 y, int* quo ); | (A) | (since C++11) (constexpr since C++23)

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 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).

(until C++23)

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, 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)

### 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)ldivlldiv (C++11) | computes quotient and remainder of integer division (function)
fmodfmodffmodl (C++11)(C++11) | remainder of the floating point division operation (function)
remainderremainderfremainderl (C++11)(C++11)(C++11) | signed remainder of the division operation (function)
C documentation for remquo

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