std::fmod, std::fmodf, std::fmodl - cppreference.com (original) (raw)

Defined in header
(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
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 - 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.

[edit] Parameters

x, y	-	floating-point or integer 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 std::fmod behaves as if implemented as follows:

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)

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

#include #include #include // #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: