std::rotl
From cppreference.com
| Defined in header <bit>
|
||
| template<class T> [[nodiscard]] constexpr T rotl(T x, int s) noexcept; |
(since C++20) | |
Computes the result of bitwise left-rotating the value of x by s positions. This operation is also known as a left circular shift.
Formally, let N be std::numeric_limits<T>::digits, r be s % N.
- If
ris 0, returns x; - if
ris positive, returns (x << r) | (x >> (N - r)); - if
ris negative, returns std::rotr(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 left-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 << "rotl(i,0) = " << std::bitset<8>(std::rotl(i,0)) << '\n'; std::cout << "rotl(i,1) = " << std::bitset<8>(std::rotl(i,1)) << '\n'; std::cout << "rotl(i,4) = " << std::bitset<8>(std::rotl(i,4)) << '\n'; std::cout << "rotl(i,9) = " << std::bitset<8>(std::rotl(i,9)) << '\n'; std::cout << "rotl(i,-1) = " << std::bitset<8>(std::rotl(i,-1)) << '\n'; }
Output:
i = 00011101 rotl(i,0) = 00011101 rotl(i,1) = 00111010 rotl(i,4) = 11010001 rotl(i,9) = 00111010 rotl(i,-1) = 10001110
See also
| (C++20) |
computes the result of bitwise right-rotation (function template) |