Concatenating two std::vectors

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How do I concatenate two std::vectors?

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北海茫月

2020-11-22 12:18
You should use vector::insert
```
v1.insert(v1.end(), v2.begin(), v2.end());
```
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面向向阳花

2020-11-22 12:22

I've implemented this function which concatenates any number of containers, moving from rvalue-references and copying otherwise

namespace internal {

// Implementation detail of Concatenate, appends to a pre-reserved vector, copying or moving if
// appropriate
template<typename Target, typename Head, typename... Tail>
void AppendNoReserve(Target* target, Head&& head, Tail&&... tail) {
    // Currently, require each homogenous inputs. If there is demand, we could probably implement a
    // version that outputs a vector whose value_type is the common_type of all the containers
    // passed to it, and call it ConvertingConcatenate.
    static_assert(
            std::is_same_v<
                    typename std::decay_t<Target>::value_type,
                    typename std::decay_t<Head>::value_type>,
            "Concatenate requires each container passed to it to have the same value_type");
    if constexpr (std::is_lvalue_reference_v<Head>) {
        std::copy(head.begin(), head.end(), std::back_inserter(*target));
    } else {
        std::move(head.begin(), head.end(), std::back_inserter(*target));
    }
    if constexpr (sizeof...(Tail) > 0) {
        AppendNoReserve(target, std::forward<Tail>(tail)...);
    }
}

template<typename Head, typename... Tail>
size_t TotalSize(const Head& head, const Tail&... tail) {
    if constexpr (sizeof...(Tail) > 0) {
        return head.size() + TotalSize(tail...);
    } else {
        return head.size();
    }
}

}  // namespace internal

/// Concatenate the provided containers into a single vector. Moves from rvalue references, copies
/// otherwise.
template<typename Head, typename... Tail>
auto Concatenate(Head&& head, Tail&&... tail) {
    size_t totalSize = internal::TotalSize(head, tail...);
    std::vector<typename std::decay_t<Head>::value_type> result;
    result.reserve(totalSize);
    internal::AppendNoReserve(&result, std::forward<Head>(head), std::forward<Tail>(tail)...);
    return result;
}

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栀梦

2020-11-22 12:24

std::vector<int> first;
std::vector<int> second;

first.insert(first.end(), second.begin(), second.end());

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一整个雨季

2020-11-22 12:24

Here's a general purpose solution using C++11 move semantics:

template <typename T>
std::vector<T> concat(const std::vector<T>& lhs, const std::vector<T>& rhs)
{
    if (lhs.empty()) return rhs;
    if (rhs.empty()) return lhs;
    std::vector<T> result {};
    result.reserve(lhs.size() + rhs.size());
    result.insert(result.cend(), lhs.cbegin(), lhs.cend());
    result.insert(result.cend(), rhs.cbegin(), rhs.cend());
    return result;
}

template <typename T>
std::vector<T> concat(std::vector<T>&& lhs, const std::vector<T>& rhs)
{
    lhs.insert(lhs.cend(), rhs.cbegin(), rhs.cend());
    return std::move(lhs);
}

template <typename T>
std::vector<T> concat(const std::vector<T>& lhs, std::vector<T>&& rhs)
{
    rhs.insert(rhs.cbegin(), lhs.cbegin(), lhs.cend());
    return std::move(rhs);
}

template <typename T>
std::vector<T> concat(std::vector<T>&& lhs, std::vector<T>&& rhs)
{
    if (lhs.empty()) return std::move(rhs);
    lhs.insert(lhs.cend(), std::make_move_iterator(rhs.begin()), std::make_move_iterator(rhs.end()));
    return std::move(lhs);
}

Note how this differs from appending to a vector.

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日久生厌

2020-11-22 12:26
If what you're looking for is a way to append a vector to another after creation, vector::insert is your best bet, as has been answered several times, for example:
```
vector<int> first = {13};
const vector<int> second = {42};

first.insert(first.end(), second.cbegin(), second.cend());
```
Sadly there's no way to construct a const vector<int>, as above you must construct and then insert.

If what you're actually looking for is a container to hold the concatenation of these two vector<int>s, there may be something better available to you, if:
1. Your vector contains primitives
2. Your contained primitives are of size 32-bit or smaller
3. You want a const container
If the above are all true, I'd suggest using the basic_string who's char_type matches the size of the primitive contained in your vector. You should include a static_assert in your code to validate these sizes stay consistent:
```
static_assert(sizeof(char32_t) == sizeof(int));
```
With this holding true you can just do:
```
const u32string concatenation = u32string(first.cbegin(), first.cend()) + u32string(second.cbegin(), second.cend());
```
For more information on the differences between string and vector you can look here: https://stackoverflow.com/a/35558008/2642059

For a live example of this code you can look here: http://ideone.com/7Iww3I
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心在旅途

2020-11-22 12:28

If you are using C++11, and wish to move the elements rather than merely copying them, you can use std::move_iterator along with insert (or copy):

#include <vector>
#include <iostream>
#include <iterator>

int main(int argc, char** argv) {
  std::vector<int> dest{1,2,3,4,5};
  std::vector<int> src{6,7,8,9,10};

  // Move elements from src to dest.
  // src is left in undefined but safe-to-destruct state.
  dest.insert(
      dest.end(),
      std::make_move_iterator(src.begin()),
      std::make_move_iterator(src.end())
    );

  // Print out concatenated vector.
  std::copy(
      dest.begin(),
      dest.end(),
      std::ostream_iterator<int>(std::cout, "\n")
    );

  return 0;
}

This will not be more efficient for the example with ints, since moving them is no more efficient than copying them, but for a data structure with optimized moves, it can avoid copying unnecessary state:

#include <vector>
#include <iostream>
#include <iterator>

int main(int argc, char** argv) {
  std::vector<std::vector<int>> dest{{1,2,3,4,5}, {3,4}};
  std::vector<std::vector<int>> src{{6,7,8,9,10}};

  // Move elements from src to dest.
  // src is left in undefined but safe-to-destruct state.
  dest.insert(
      dest.end(),
      std::make_move_iterator(src.begin()),
      std::make_move_iterator(src.end())
    );

  return 0;
}

After the move, src's element is left in an undefined but safe-to-destruct state, and its former elements were transfered directly to dest's new element at the end.

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