std::rotr

 

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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).

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