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

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).


[edit] Parameters

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

[edit] Return value

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

[edit] Notes

Feature testing macro: __cpp_lib_bitops

[edit] Example

#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';


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

[edit] See also

computes the result of bitwise left-rotation
(function template) [edit]
performs binary shift left and shift right
(public member function of std::bitset<N>) [edit]