std::rotr
From cppreference.com
Defined in header <bit>
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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 only participates in overload resolution 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 | - | values of unsigned integer type |
Return value
The result of bitwise right-rotating x
by s
positions.
Example
Run this code
#include <bit> #include <bitset> #include <cstdint> #include <iostream> int main() { 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
(C++20) |
computes the result of bitwise left-rotation (function template) |