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std::rotl(3) C++ Standard Libary std::rotl(3) NAME std::rotl - std::rotl Synopsis Defined in header <bit> template< class T > (since C++20) [[nodiscard]] constexpr T rotl( T x, int s ) noexcept; Computes the result of bitwise left-rotating the value of x by s po- sitions. This operation is also known as a left 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::rotr(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 left-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 << "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 rotr computes the result of bitwise right-rotation (C++20) (function template) operator<<= operator>>= performs binary shift left and shift right operator<< (public member function of std::bitset<N>) operator>> http://cppreference.com 2022.07.31 std::rotl(3)
NAME | Synopsis | Parameters | Return value | Notes | Example | Output: | See also
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