Namespaces
Variants
Actions

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

From cppreference.com
< cpp‎ | numeric‎ | math
 
 
 
 
Defined in header <cmath>
(1)
float       fmod (float x, float y );

double      fmod (double x, double y );

longdouble fmod (longdouble x, longdouble 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)
longdouble fmodl(longdouble x, longdouble y );
(3)(since C++11)
(constexpr since C++23)
SIMD overload(since C++26)
Defined in header <simd>
template<class V0, class V1 >

constexpr/*math-common-simd-t*/<V0, V1>

            fmod (const V0& v_x, const V1& v_y );
(S) (since C++26)
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)
S) The SIMD overload performs an element-wise std::fmod on v_xand v_y.
(See math-common-simd-t for its definition.)
(since C++26)
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.

Contents

[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 [-0.065535.0], which corresponds to unsignedshort, but std::remainder(std::rint(x), 65536.0 is in the range [-32767.0+32768.0], which is outside of the range of signedshort.

The double version of std::fmod behaves as if implemented as follows:

double fmod(double x, double y){#pragma STDC FENV_ACCESS ONdouble result =std::remainder(std::fabs(x), y =std::fabs(y));if(std::signbit(result)) result += y;returnstd::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 longdouble, then std::fmod(num1, num2) has the same effect as std::fmod(static_cast<longdouble>(num1),
              static_cast<longdouble>(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 <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 valuesstd::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 handlingstd::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

[edit]See also

computes quotient and remainder of integer division
(function)[edit]
(C++11)(C++11)(C++11)
signed remainder of the division operation
(function)[edit]
(C++11)(C++11)(C++11)
signed remainder as well as the three last bits of the division operation
(function)[edit]
close