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

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< cpp‎ | numeric‎ | math

Defined in header `<cmath>` |  |
| (1) |
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 ); | (2) | (since C++11) (constexpr since C++23)
long double fmodl( long double x, long double y ); | (3) | (since C++11) (constexpr since C++23)
Additional overloads (since C++11) |  |
Defined in header `<cmath>` |  |
template< class Integer > double fmod ( Integer x, Integer y ); | (A) | (constexpr since C++23)

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)

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 - 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.

### 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

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),

- 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)). | (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)

### 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

### See also

div(int)ldiv lldiv (C++11) | computes quotient and remainder of integer division (function)
remainder remainderf remainderl (C++11)(C++11)(C++11) | signed remainder of the division operation (function)
remquo remquof remquol (C++11)(C++11)(C++11) | signed remainder as well as the three last bits of the division operation (function)

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