'Concatenating two std::vectors
How do I concatenate two std::vectors?
Solution 1:[1]
vector1.insert( vector1.end(), vector2.begin(), vector2.end() );
Solution 2:[2]
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.
Solution 3:[3]
I would use the insert function, something like:
vector<int> a, b;
//fill with data
b.insert(b.end(), a.begin(), a.end());
Solution 4:[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.
Solution 5:[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>.
Solution 6:[6]
std::vector<int> first;
std::vector<int> second;
first.insert(first.end(), second.begin(), second.end());
Solution 7:[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());
}
Solution 8:[8]
Solution 9:[9]
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 {
v2.insert(v2.end(),v1.begin(),v1.end());
}
Solution 10:[10]
If you want to be able to concatenate vectors concisely, you could overload the += operator.
template <typename T>
std::vector<T>& operator +=(std::vector<T>& vector1, const std::vector<T>& vector2) {
vector1.insert(vector1.end(), vector2.begin(), vector2.end());
return vector1;
}
Then you can call it like this:
vector1 += vector2;
Solution 11:[11]
There is an algorithm std::merge from C++17, which is very easy to use when the input vectors are sorted,
Below is the example:
#include <iostream>
#include <vector>
#include <algorithm>
int main()
{
//DATA
std::vector<int> v1{2,4,6,8};
std::vector<int> v2{12,14,16,18};
//MERGE
std::vector<int> dst;
std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(dst));
//PRINT
for(auto item:dst)
std::cout<<item<<" ";
return 0;
}
Solution 12:[12]
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).
Solution 13:[13]
You should use vector::insert
v1.insert(v1.end(), v2.begin(), v2.end());
Solution 14:[14]
If your goal is simply to iterate over the range of values for read-only purposes, an alternative is to wrap both vectors around a proxy (O(1)) instead of copying them (O(n)), so they are promptly seen as a single, contiguous one.
std::vector<int> A{ 1, 2, 3, 4, 5};
std::vector<int> B{ 10, 20, 30 };
VecProxy<int> AB(A, B); // ----> O(1)!
for (size_t i = 0; i < AB.size(); i++)
std::cout << AB[i] << " "; // ----> 1 2 3 4 5 10 20 30
Refer to https://stackoverflow.com/a/55838758/2379625 for more details, including the 'VecProxy' implementation as well as pros & cons.
Solution 15:[15]
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]
Solution 16:[16]
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.
Solution 17:[17]
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
Solution 18:[18]
vector<int> v1 = {1, 2, 3, 4, 5};
vector<int> v2 = {11, 12, 13, 14, 15};
copy(v2.begin(), v2.end(), back_inserter(v1));
Solution 19:[19]
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;
}
Solution 20:[20]
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.
Solution 21:[21]
For containers which offer push_back (string, vector, deque, ...):
std::copy(std::begin(input), std::end(input), std::back_inserter(output))
and
for containers which offer insert (maps, sets):
std::copy(std::begin(input), std::end(input), std::inserter(output, output.end()))
Solution 22:[22]
Using C++20 you can get rid of begin() and end() with ranges.
#include <ranges>
std::ranges::copy(vec2, std::back_inserter(vec1));
or if you want to move elements:
std::ranges::move(vec2, std::back_inserter(vec1));
Solution 23:[23]
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:
- Your
vectorcontains primitives - Your contained primitives are of size 32-bit or smaller
- You want a
constcontainer
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
Solution 24:[24]
You can do it with pre-implemented STL algorithms using a template for a polymorphic type use.
#include <iostream>
#include <vector>
#include <algorithm>
template<typename T>
void concat(std::vector<T>& valuesa, std::vector<T>& valuesb){
for_each(valuesb.begin(), valuesb.end(), [&](int value){ valuesa.push_back(value);});
}
int main()
{
std::vector<int> values_p={1,2,3,4,5};
std::vector<int> values_s={6,7};
concat(values_p, values_s);
for(auto& it : values_p){
std::cout<<it<<std::endl;
}
return 0;
}
You can clear the second vector if you don't want to use it further (clear() method).
Solution 25:[25]
Concatenate two std::vector-s with for loop in one std::vector.
std::vector <int> v1 {1, 2, 3}; //declare vector1
std::vector <int> v2 {4, 5}; //declare vector2
std::vector <int> suma; //declare vector suma
for(int i = 0; i < v1.size(); i++) //for loop 1
{
suma.push_back(v1[i]);
}
for(int i = 0; i< v2.size(); i++) //for loop 2
{
suma.push_back(v2[i]);
}
for(int i = 0; i < suma.size(); i++) //for loop 3-output
{
std::cout << suma[i];
}
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