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# std::fmod

From cppreference.com
< cpp‎ | numeric‎ | math

Defined in header `<cmath>` |  |
float fmod( float x, float y ); | (1) |
double fmod( double x, double y ); | (2) |
long double fmod( long double x, long double y ); | (3) |
Promoted fmod( Arithmetic1 x, Arithmetic2 y ); | (4) | (since C++11)

1-3) Computes the floating-point remainder of the division operation x/y.
4) A set of overloads or a function template for all combinations of arguments of arithmetic type not covered by 1-3. If any argument has integral type, it is cast to double. If any other argument is long double, then the return type is long double, otherwise it is double.

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

The returned value has the same sign as `x` and is less or equal to `y` in magnitude.

### [edit] Parameters

x, y | - | floating point values

### [edit] 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.

### [edit] 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

### [edit] 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 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);
    }

### [edit] Example

Run this code

    #include <iostream>
    #include <cmath>
    #include <cfenv>

    #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(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

### [edit] See also

div(int)ldivlldiv (C++11) | computes quotient and remainder of integer division (function) [edit]
remainder (C++11) | signed remainder of the division operation (function) [edit]
remquo (C++11) | signed remainder as well as the three last bits of the division operation (function) [edit]
C documentation for fmod

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