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std::rotr(3) C++ Standard Libary std::rotr(3)

NAME

std::rotr - std::rotr

Synopsis


Defined in header <bit>
template< class T > (since C++20)
[[nodiscard]] constexpr T rotr( T x, int s ) noexcept;


Computes the result of bitwise right-rotating the value of x by s positions. This
operation is also known as a right circular shift.


Formally, let N be std::numeric_limits<T>::digits, r be s % N.


* If r is 0, returns x;
* if r is positive, returns (x >> r) | (x << (N - r));
* if r is negative, returns std::rotl(x, -r).


This overload participates in overload resolution only if T is an unsigned integer
type (that is, unsigned char, unsigned short, unsigned int, unsigned long, unsigned
long long, or an extended unsigned integer type).

Parameters


x - value of unsigned integer type
s - number of positions to shift

Return value


The result of bitwise right-rotating x by s positions.

Notes


Feature-test macro: __cpp_lib_bitops

Example

// Run this code


#include <bit>
#include <bitset>
#include <cstdint>
#include <iostream>


int main()
{
const std::uint8_t i = 0b00011101;
std::cout << "i = " << std::bitset<8>(i) << '\n';
std::cout << "rotr(i,0) = " << std::bitset<8>(std::rotr(i,0)) << '\n';
std::cout << "rotr(i,1) = " << std::bitset<8>(std::rotr(i,1)) << '\n';
std::cout << "rotr(i,9) = " << std::bitset<8>(std::rotr(i,9)) << '\n';
std::cout << "rotr(i,-1) = " << std::bitset<8>(std::rotr(i,-1)) << '\n';
}

Output:


i = 00011101
rotr(i,0) = 00011101
rotr(i,1) = 10001110
rotr(i,9) = 10001110
rotr(i,-1) = 00111010

See also


rotl computes the result of bitwise left-rotation
(C++20) (function template)
operator<<=
operator>>= performs binary shift left and shift right
operator<< (public member function of std::bitset<N>)
operator>>

2022.07.31 http://cppreference.com