std::rotr
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, returnsx
; - if
r
is positive, returns(x >> r) | (x << (N - r))
; - if
r
is negative, returnsstd::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 testing macro: __cpp_lib_bitops
.
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'; }
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) |
performs binary shift left and shift right (public member function of std::bitset<N> ) |
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