在Bradgonesurfing的答案中，很多时候并不需要连接两个向量(O(n))，而是像连接(O(1))一样处理它们。如果这是您的情况，可以在不需要Boost库的情况下完成。

诀窍是创建一个向量代理:一个包装器类，它操纵对两个向量的引用，在外部被视为单个连续的向量。

使用

std::vector<int> A{ 1, 2, 3, 4, 5};
std::vector<int> B{ 10, 20, 30 };

VecProxy<int> AB(A, B);  // ----> O(1). No copies performed.

for (size_t i = 0; i < AB.size(); ++i)
    std::cout << AB[i] << " ";  // 1 2 3 4 5 10 20 30

实现

template <class T>
class VecProxy {
private:
    std::vector<T>& v1, v2;
public:
    VecProxy(std::vector<T>& ref1, std::vector<T>& ref2) : v1(ref1), v2(ref2) {}
    const T& operator[](const size_t& i) const;
    const size_t size() const;
};

template <class T>
const T& VecProxy<T>::operator[](const size_t& i) const{
    return (i < v1.size()) ? v1[i] : v2[i - v1.size()];
};

template <class T>
const size_t VecProxy<T>::size() const { return v1.size() + v2.size(); };

主要好处

它是O(1)(常数时间)来创建它，并且使用最小的额外内存分配。

一些需要考虑的问题

You should only go for it if you really know what you're doing when dealing with references. This solution is intended for the specific purpose of the question made, for which it works pretty well. To employ it in any other context may lead to unexpected behavior if you are not sure on how references work. In this example, AB does not provide a non-const access operator ([ ]). Feel free to include it, but keep in mind: since AB contains references, to assign it values will also affect the original elements within A and/or B. Whether or not this is a desirable feature, it's an application-specific question one should carefully consider. Any changes directly made to either A or B (like assigning values, sorting, etc.) will also "modify" AB. This is not necessarily bad (actually, it can be very handy: AB does never need to be explicitly updated to keep itself synchronized to both A and B), but it's certainly a behavior one must be aware of. Important exception: to resize A and/or B to sth bigger may lead these to be reallocated in memory (for the need of contiguous space), and this would in turn invalidate AB. Because every access to an element is preceded by a test (namely, "i < v1.size()"), VecProxy access time, although constant, is also a bit slower than that of vectors. This approach can be generalized to n vectors. I haven't tried, but it shouldn't be a big deal.

这取决于你是否真的需要物理上连接两个向量或者你想为了迭代而给出连接的外观。boost::join函数

http://www.boost.org/doc/libs/1_43_0/libs/range/doc/html/range/reference/utilities/join.html

会给你这个。

std::vector<int> v0;
v0.push_back(1);
v0.push_back(2);
v0.push_back(3);

std::vector<int> v1;
v1.push_back(4);
v1.push_back(5);
v1.push_back(6);
...

BOOST_FOREACH(const int & i, boost::join(v0, v1)){
    cout << i << endl;
}

应该给你

1
2
3
4
5
6

注意boost::join不会将两个向量复制到一个新的容器中 但生成一对覆盖的跨度的迭代器(range) 这两个容器。会有一些性能开销，但也许 而不是先将所有数据复制到一个新容器中。

这正是成员函数std::vector::insert的作用

std::vector<int> AB = A;
AB.insert(AB.end(), B.begin(), B.end());

所有的解决方案都是正确的，但我发现写一个函数来实现它更容易。是这样的:

template <class T1, class T2>
void ContainerInsert(T1 t1, T2 t2)
{
    t1->insert(t1->end(), t2->begin(), t2->end());
}

这样你就可以避免像这样的临时放置:

ContainerInsert(vec, GetSomeVector());

AB.reserve( A.size() + B.size() ); // preallocate memory
AB.insert( AB.end(), A.begin(), A.end() );
AB.insert( AB.end(), B.begin(), B.end() );

基于Kiril V. Lyadvinsky的回答，我做了一个新的版本。这个片段使用模板和重载。有了它，你可以写vector3 = vector1 + vector2和vector4 += vector3。希望能有所帮助。

template <typename T>
std::vector<T> operator+(const std::vector<T> &A, const std::vector<T> &B)
{
    std::vector<T> AB;
    AB.reserve(A.size() + B.size());                // preallocate memory
    AB.insert(AB.end(), A.begin(), A.end());        // add A;
    AB.insert(AB.end(), B.begin(), B.end());        // add B;
    return AB;
}

template <typename T>
std::vector<T> &operator+=(std::vector<T> &A, const std::vector<T> &B)
{
    A.reserve(A.size() + B.size());                // preallocate memory without erase original data
    A.insert(A.end(), B.begin(), B.end());         // add B;
    return A;                                        // here A could be named AB
}