# copy_backward

Category: algorithms | Component type: function

### Prototype

    template <class BidirectionalIterator1, class BidirectionalIterator2>
    BidirectionalIterator2 copy_backward(BidirectionalIterator1 first,
                                         BidirectionalIterator1 last,
                                         BidirectionalIterator2 result);

### Description

`Copy_backward` copies elements from the range `[first, last)` to the range `[result - (last - first), result)` [1]. That is, it performs the assignments `*(result - 1) = *(last - 1)`, `*(result - 2) = *(last - 2)`, and so on. Generally, for every integer `n` from `0` to `last - first`, `copy_backward` performs the assignment `*(result - n - 1) = *(last - n - 1)`. Assignments are performed from the end of the input sequence to the beginning, i.e. in order of increasing `n`. [2]

The return value is `result - (last - first)`

### Definition

Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h.

### Requirements on types

- BidirectionalIterator1 and BidirectionalIterator2 are models of BidirectionalIterator.
- BidirectionalIterator1's value type is convertible to BidirectionalIterator2's value type.

### Preconditions

- `[first, last)` is a valid range.
- `result` is not an iterator within the range `[first, last)`.
- There is enough space to hold all of the elements being copied. More formally, the requirement is that `[result - (last - first), result)` is a valid range.

### Complexity

Linear. Exactly `last - first` assignments are performed.

### Example

    vector<int> V(15);
    iota(V.begin(), V.end(), 1);
    copy_backward(V.begin(), V.begin() + 10, V.begin() + 15);

### Notes

[1] `Result` is an iterator that points to the end of the output range. This is highly unusual: in all other STL algorithms that denote an output range by a single iterator, that iterator points to the beginning of the range.

[2] The order of assignments matters in the case where the input and output ranges overlap: `copy_backward` may not be used if `result` is in the range `[first, last)`. That is, it may not be used if the end of the output range overlaps with the input range, but it may be used if the beginning of the output range overlaps with the input range; `copy` has opposite restrictions. If the two ranges are completely nonoverlapping, of course, then either algorithm may be used.

### See also

`copy`, `copy_n`

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