std::remainder, std::remainderf, std::remainderl - cppreference.com

Defined in header

<cmath>

floatremainder(floatx,floaty);doubleremainder(doublex,doubley);longdoubleremainder(longdoublex,longdoubley); (1)(until C++23)constexpr/*floating-point-type*/remainder(/*floating-point-type*/x,/*floating-point-type*/y);(since C++23)floatremainderf(floatx,floaty); (2)(since C++11)
(constexpr since C++23)longdoubleremainderl(longdoublex,longdoubley); (3)(since C++11)
(constexpr since C++23)

SIMD overload

(since C++26)

Defined in header

<simd>

template<classV0,classV1>constexpr/*math-common-simd-t*/<V0,V1>remainder(constV0&v_x,constV1&v_y); (S) (since C++26)

Additional overloads

(since C++11)

Defined in header

<cmath>

template<classInteger>doubleremainder(Integerx,Integery); (A)(constexpr since C++23)1-3) Computes the IEEE remainder of the floating point division operation x/y. The library provides overloads of std::remainder 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::remainder 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 IEEE floating-point remainder of the division operation x/y calculated by this function is exactly the value x-quo*y, where the value quo is the integral value nearest the exact value x/y. When |quo - x / y| = ½, the value quo is chosen to be even.

In contrast to

std::fmod

, the returned value is not guaranteed to have the same sign as x.

If the returned value is zero, it will have the same sign as x.

Parameters

x, y - floating-point or integer values Return value

If successful, returns the IEEE 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 is returned.

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, the result is always exact.

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 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 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 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::remainder(num1,num2) has the same effect as std::remainder(static_cast<longdouble>(num1),
static_cast<longdouble>(num2)).

Otherwise, if num1 and/or num2 has type double or an integer type, then std::remainder(num1,num2) has the same effect as std::remainder(static_cast<double>(num1),
static_cast<double>(num2)).

Otherwise, if num1 or num2 has type float, then std::remainder(num1,num2) has the same effect as std::remainder(static_cast<float>(num1),
static_cast<float>(num2)).

(until C++23)If num1 and num2 have arithmetic types, then std::remainder(num1,num2) has the same effect as std::remainder(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 ONintmain(){std::cout<<"remainder(+5.1, +3.0) = "<<std::remainder(5.1,3)<<'\n'<<"remainder(-5.1, +3.0) = "<<std::remainder(-5.1,3)<<'\n'<<"remainder(+5.1, -3.0) = "<<std::remainder(5.1,-3)<<'\n'<<"remainder(-5.1, -3.0) = "<<std::remainder(-5.1,-3)<<'\n';// special valuesstd::cout<<"remainder(-0.0, 1.0) = "<<std::remainder(-0.0,1)<<'\n'<<"remainder(5.1, Inf) = "<<std::remainder(5.1,INFINITY)<<'\n';// error handlingstd::feclearexcept(FE_ALL_EXCEPT);std::cout<<"remainder(+5.1, 0) = "<<std::remainder(5.1,0)<<'\n';if(fetestexcept(FE_INVALID))std::cout<<" FE_INVALID raised\n";}Possible output:

remainder(+5.1, +3.0) = -0.9 remainder(-5.1, +3.0) = 0.9 remainder(+5.1, -3.0) = -0.9 remainder(-5.1, -3.0) = 0.9 remainder(-0.0, 1.0) = -0 remainder(5.1, Inf) = 5.1 remainder(+5.1, 0) = -nan FE_INVALID raised See also