std::shift_...shift_right(3)

💅 std::shift_...shift_right(3) 💅

Bestie, welcome to the C++ Standard Libary manual page for std::shift_left and std::shift_right. Yes, that's the manual name, we're not fixing it, it's iconic. Buckle up: these algorithms slide a range left or right, and the details are lowkey doing the most.

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✨ NAME ✨

std::shift_left,std::shift_right — std::shift_left,std::shift_right

That's the NAME section, and yes it's giving mirror selfie. Two algorithms, one shift-y family. Periodt.

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🌸 Synopsis 🌸

Defined in header <algorithm>.

template< class ForwardIt >
constexpr ForwardIt shift_left( ForwardIt first, ForwardIt last,
                                typename std::iterator_traits<ForwardIt>::difference_type n );

(1) — (since C++20)

template< class ExecutionPolicy, class ForwardIt >
ForwardIt shift_left( ExecutionPolicy&& policy,
                      ForwardIt first, ForwardIt last,
                      typename std::iterator_traits<ForwardIt>::difference_type n );

(2) — (since C++20)

template< class ForwardIt >
constexpr ForwardIt shift_right( ForwardIt first, ForwardIt last,
                                 typename std::iterator_traits<ForwardIt>::difference_type n );

(3) — (since C++20)

template< class ExecutionPolicy, class ForwardIt >
ForwardIt shift_right( ExecutionPolicy&& policy,
                       ForwardIt first, ForwardIt last,
                       typename std::iterator_traits<ForwardIt>::difference_type n );

(4) — (since C++20)

Shifts the elements in the range [first, last) by n positions.

• (1) — shifts the elements toward the beginning of the range.
  • If n == 0 || n >= last - first, there are no effects.
  • Otherwise, for every integer i in [0, last - first - n), moves the element originally at position first + n + i to position first + i. The moves are performed in increasing order of i starting from 0.
• (3) — shifts the elements toward the end of the range.
  • If n == 0 || n >= last - first, there are no effects.
  • Otherwise, for every integer i in [0, last - first - n), moves the element originally at position first + i to position first + n + i. If ForwardIt meets the LegacyBidirectionalIterator requirements, then the moves are performed in decreasing order of i starting from last - first - n - 1.
• (2), (4) — same as (1) and (3), respectively, but executed according to policy and the moves may be performed in any order.

These overloads participate in overload resolution only if std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>> is true.

Elements that are in the original range but not the new range are left in a valid but unspecified state. So like, they're alive, but they're not telling you what they've been through.

🚨 If any of the following conditions is satisfied, the behavior is undefined: - n >= 0 is not true. - The type of *first is not MoveAssignable. - For shift_right, ForwardIt is neither LegacyBidirectionalIterator nor ValueSwappable.

That's the danger zone, bestie. Do not.

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🦋 Parameters 🦋

• first — the beginning of the original range.
• last — the end of the original range.
• n — the number of positions to shift.
• policy — the execution policy to use. See execution policy for details.

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👑 Type requirements 👑

• ForwardIt must meet the requirements of LegacyForwardIterator.

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🎀 Return value 🎀

• (1), (2) — the end of the resulting range.
  • If n is less than std::distance(first, last), returns an iterator equal to std::next(first, (std::distance(first, last) - n)).
  • Otherwise, returns first.
• (3), (4) — the beginning of the resulting range.
  • If n is less than std::distance(first, last), returns an iterator equal to std::next(first, n).
  • Otherwise, returns last.

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💀 Complexity 💀

• (1), (2) — At most std::distance(first, last) - n assignments.
• (3), (4) — At most std::distance(first, last) - n assignment or swaps. Yes, that's the wording. We're keeping it.

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🫶 Exceptions 🫶

The overloads with a template parameter named ExecutionPolicy report errors as follows:

• If execution of a function invoked as part of the algorithm throws an exception and ExecutionPolicy is one of the standard policies, std::terminate is called. For any other ExecutionPolicy, the behavior is implementation-defined.
• If the algorithm fails to allocate memory, std::bad_alloc is thrown.

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🌈 Notes 🌈

• Feature-test macro: __cpp_lib_shift — Value: 201806L — Std: (C++20) — Feature: std::shift_left and std::shift_right.

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🍓 Example 🍓

// Run this code

#include <algorithm>
#include <iostream>
#include <string>
#include <type_traits>
#include <vector>

struct S
{
    int value{0};
    bool specified_state{true};

    S(int v = 0) : value{v} {}
    S(S const& rhs) = default;
    S(S&& rhs) { *this = std::move(rhs); }
    S& operator=(S const& rhs) = default;
    S& operator=(S&& rhs)
    {
        if (this != &rhs)
        {
            value = rhs.value;
            specified_state = rhs.specified_state;
            rhs.specified_state = false;
        }
        return *this;
    }
};

template<typename T>
std::ostream& operator<<(std::ostream& os, std::vector<T> const& v)
{
    for (const auto& s : v)
    {
        if constexpr (std::is_same_v<T, S>)
            s.specified_state ? os << s.value << ' ' : os << ". ";
        else if constexpr (std::is_same_v<T, std::string>)
            os << (s.empty() ? "." : s) << ' ';
        else
            os << s << ' ';
    }
    return os;
}

int main()
{
    std::cout << std::left;

    std::vector<S>           a{1, 2, 3, 4, 5, 6, 7};
    std::vector<int>         b{1, 2, 3, 4, 5, 6, 7};
    std::vector<std::string> c{"α", "β", "γ", "δ", "ε", "ζ", "η"};

    std::cout << "vector<S> \tvector<int> \tvector<string>\n";
    std::cout << a << "  " << b << "  " << c << '\n';

    std::shift_left(begin(a), end(a), 3);
    std::shift_left(begin(b), end(b), 3);
    std::shift_left(begin(c), end(c), 3);
    std::cout << a << "  " << b << "  " << c << '\n';

    std::shift_right(begin(a), end(a), 2);
    std::shift_right(begin(b), end(b), 2);
    std::shift_right(begin(c), end(c), 2);
    std::cout << a << "  " << b << "  " << c << '\n';

    std::shift_left(begin(a), end(a), 8); // has no effect: n >= last - first
    std::shift_left(begin(b), end(b), 8); // ditto
    std::shift_left(begin(c), end(c), 8); // ditto
    std::cout << a << "  " << b << "  " << c << '\n';

//  std::shift_left(begin(a), end(a), -3); // UB, e.g. segfault
}

Possible output:

vector<S>       vector<int>     vector<string>
1 2 3 4 5 6 7   1 2 3 4 5 6 7   α β γ δ ε ζ η
4 5 6 7 . . .   4 5 6 7 5 6 7   δ ε ζ η . . .
. . 4 5 6 7 .   4 5 4 5 6 7 5   . . δ ε ζ η .
. . 4 5 6 7 .   4 5 4 5 6 7 5   . . δ ε ζ η .

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⭐ See also ⭐

• move — moves a range of elements to a new location. (C++11) (function template)
• move_backward — moves a range of elements to a new location in backwards order. (C++11) (function template)
• rotate — rotates the order of elements in a range. (function template)
• ranges::shift_left, ranges::shift_right — shifts elements in a range. (C++23) (niebloid)

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Source/version sign-off: http://cppreference.com, dated 2024.06.10, page std::shift_...shift_right(3). That's the whole manual page, bestie — go shift responsibly. 💖✨