Concatenating two std::vectors

问题:

How do I concatenate two std::vectors?

回答1:

vector1.insert( vector1.end(), vector2.begin(), vector2.end() );

回答2:

If you are using C++11, and wish to move the elements rather than merely copying them, you can use std::move_iterator (http://en.cppreference.com/w/cpp/iterator/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.

回答3:

I would use the insert function, something like:

vector<int> a, b;
//fill with data
b.insert(b.end(), a.begin(), a.end());

回答4:

Or you could use:

std::copy(source.begin(), source.end(), std::back_inserter(destination));

This pattern is useful if the two vectors don\'t contain exactly the same type of thing, because you can use something instead of std::back_inserter to convert from one type to the other.

回答5:

With C++11, I\'d prefer following to append vector b to a:

std::move(b.begin(), b.end(), std::back_inserter(a));

when a and b are not overlapped, and b is not going to be used anymore.

---

This is std::move from <algorithm>, not the usual std::move from <utility>.

回答6:

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

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

回答7:

I prefer one that is already mentioned:

a.insert(a.end(), b.begin(), b.end());

But if you use C++11, there is one more generic way:

a.insert(std::end(a), std::begin(b), std::end(b));

---

Also, not part of a question, but it is advisable to use reserve before appending for better performance. And if you are concatenating vector with itself, without reserving it fails, so you always should reserve.

---

So basically what you need:

template <typename T>
void Append(std::vector<T>& a, const std::vector<T>& b)
{
    a.reserve(a.size() + b.size());
    a.insert(a.end(), b.begin(), b.end());
}

回答8:

You should use vector::insert

v1.insert(v1.end(), v2.begin(), v2.end());

回答9:

With range v3, you may have a lazy concatenation:

ranges::view::concat(v1, v2)

Demo.

回答10:

If you are interested in strong exception guarantee (when copy constructor can throw an exception):

template<typename T>
inline void append_copy(std::vector<T>& v1, const std::vector<T>& v2)
{
    const auto orig_v1_size = v1.size();
    v1.reserve(orig_v1_size + v2.size());
    try
    {
        v1.insert(v1.end(), v2.begin(), v2.end());
    }
    catch(...)
    {
        v1.erase(v1.begin() + orig_v1_size, v1.end());
        throw;
    }
}

Similar append_move with strong guarantee can\'t be implemented in general if vector element\'s move constructor can throw (which is unlikely but still).

回答11:

vector<int> v1 = {1, 2, 3, 4, 5};
vector<int> v2 = {11, 12, 13, 14, 15};
copy(v2.begin(), v2.end(), back_inserter(v1));

回答12:

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.

回答13:

Add this one to your header file:

template <typename T> vector<T> concat(vector<T> &a, vector<T> &b) {
    vector<T> ret = vector<T>();
    copy(a.begin(), a.end(), back_inserter(ret));
    copy(b.begin(), b.end(), back_inserter(ret));
    return ret;
}

and use it this way:

vector<int> a = vector<int>();
vector<int> b = vector<int>();

a.push_back(1);
a.push_back(2);
b.push_back(62);

vector<int> r = concat(a, b);

r will contain [1,2,62]

回答14:

A general performance boost for concatenate is to check the size of the vectors. And merge/insert the smaller one with the larger one.

//vector<int> v1,v2;
if(v1.size()>v2.size()){
    v1.insert(v1.end(),v2.begin(),v2.end());
}else{
    v1.insert(v2.end(),v1.begin(),v1.end());
}

回答15:

You can prepare your own template for + operator:

template <typename T> 
inline T operator+(const T & a, const T & b)
{
    T res = a;
    res.insert(res.end(), b.begin(), b.end());
    return res;
}

Next thing - just use +:

vector<int> a{1, 2, 3, 4};
vector<int> b{5, 6, 7, 8};
for (auto x: a + b)
    cout << x << \" \";
cout << endl;

This example gives output:

1 2 3 4 5 6 7 8

回答16:

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

回答17:

To be honest, you could fast concatenate two vectors by copy elements from two vectors into the other one or just only append one of two vectors!. It depends on your aim.

Method 1: Assign new vector with its size is the sum of two original vectors\' size.

vector<int> concat_vector = vector<int>();
concat_vector.setcapacity(vector_A.size() + vector_B.size());
// Loop for copy elements in two vectors into concat_vector

Method 2: Append vector A by adding/inserting elements of vector B.

// Loop for insert elements of vector_B into vector_A with insert() 
function: vector_A.insert(vector_A .end(), vector_B.cbegin(), vector_B.cend());

回答18:

This solution might be a bit complicated, but boost-range has also some other nice things to offer.

#include <iostream>
#include <vector>
#include <boost/range/algorithm/copy.hpp>

int main(int, char**) {
    std::vector<int> a = { 1,2,3 };
    std::vector<int> b = { 4,5,6 };
    boost::copy(b, std::back_inserter(a));
    for (auto& iter : a) {
        std::cout << iter << \" \";
    }
    return EXIT_SUCCESS;
}

Often ones intention is to combine vector a and b just iterate over it doing some operation. In this case, there is the ridiculous simple join function.

#include <iostream>
#include <vector>
#include <boost/range/join.hpp>
#include <boost/range/algorithm/copy.hpp>

int main(int, char**) {
    std::vector<int> a = { 1,2,3 };
    std::vector<int> b = { 4,5,6 };
    std::vector<int> c = { 7,8,9 };
    // Just creates an iterator
    for (auto& iter : boost::join(a, boost::join(b, c))) {
        std::cout << iter << \" \";
    }
    std::cout << \"\\n\";
    // Can also be used to create a copy
    std::vector<int> d;
    boost::copy(boost::join(a, boost::join(b, c)), std::back_inserter(d));
    for (auto& iter : d) {
        std::cout << iter << \" \";
    }
    return EXIT_SUCCESS;
}

For large vectors this might be an advantage, as there is no copying. It can be also used for copying an generalizes easily to more than one container.

For some reason there is nothing like boost::join(a,b,c), which could be reasonable.