# std::rotr

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< cpp‎ | numeric

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 | Value | Std | Comment
--- | --- | --- | ---
`__cpp_lib_bitops` | 201907L | (C++20) | Bit operations

### 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>`)

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