# std::ranges::shift_left, std::ranges::shift_right

< cpp‎ | algorithm‎ | ranges
 Defined in header  Call signature template< std::permutable I, std::sentinel_for S > constexpr ranges::subrange     shift_left( I first, S last, std::iter_difference_t n ); (1) (since C++23) template< ranges::forward_range R > requires std::permutable> constexpr ranges::borrowed_subrange_t     shift_left( R&& r, ranges::range_difference_t n ); (2) (since C++23) template< std::permutable I, std::sentinel_for S > constexpr ranges::subrange     shift_right( I first, S last, std::iter_difference_t n ); (3) (since C++23) template< ranges::forward_range R > requires std::permutable> constexpr ranges::borrowed_subrange_t     shift_right( R&& r, ranges::range_difference_t n ); (4) (since C++23)

Shifts the elements in the range [first, last) or r by n positions. The behavior is undefined if [first, last) is not a valid range.

1) Shifts the elements towards the beginning of the range:
• If n == 0 || n >= last - first, there are no effects.
• If n < 0, the behavior is undefined.
• 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 towards the end of the range:
• If n == 0 || n >= last - first, there are no effects.
• If n < 0, the behavior is undefined.
• Otherwise, for every integer i in [0, last - first - n), moves the element originally at position first + i to position first + n + i. If I models bidirectional_iterator, then the moves are performed in decreasing order of i starting from last - first - n - 1.
2, 4) Same as (1) or (3) respectively, but uses r as the range, as if using as first and ranges::end(r) as last.

Elements that are in the original range but not the new range are left in a valid but unspecified state.

In practice, they may be implemented as function objects, or with special compiler extensions.

## Contents

### Parameters

 first - the beginning of the original range last - the end of the original range r - the range of elements to shift n - the number of positions to shift

### Return value

1-2) {first, /*NEW_LAST*/}, where NEW_LAST is the end of the resulting range and equivalent to:
• first + (last - first - n), if n is less than last - first;
• first otherwise.
3-4) {/*NEW_FIRST*/, last}, where NEW_FIRST is the beginning of the resulting range and equivalent to:
• first + n, if n is less than last - first;
• last otherwise.

###  Complexity

1-2) At most ranges::distance(first, last) - n assignments.
3-4) At most ranges::distance(first, last) - n assignment or swaps.

### Notes

ranges::shift_left / ranges::shift_right has better efficiency on common implementations if I models bidirectional_iterator or (better) random_access_iterator.

Implementations (e.g. MSVC STL) may enable vectorization when the iterator type models contiguous_iterator and swapping its value type calls neither non-trivial special member function nor ADL-found swap.

 Feature testing macro: __cpp_lib_shift

### Example

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

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

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

//  std::ranges::shift_left(a, -3);  // UB
}

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 .