我如何连接两个std::向量?
当前回答
我将使用插入函数，类似于:
vector<int> a, b;
//fill with data
b.insert(b.end(), a.begin(), a.end());
其他回答
c++ 17中有一个算法std::merge，当输入向量排序时，它非常容易使用，
下面是例子:
#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;
}
如果希望能够简洁地连接向量，可以重载+=运算符。
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;
}
然后你可以这样调用它:
vector1 += vector2;
我已经实现了这个函数，它连接任何数量的容器，从右值引用移动和复制
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;
}
在c++ 11中，我更喜欢将向量b附加到a:
std::move(b.begin(), b.end(), std::back_inserter(a));
当a和b不重叠时，b不会再被用到。
这是std::move from <algorithm>，而不是通常的std::move from <utility>。
如果你对强异常保证感兴趣(当复制构造函数可以抛出异常时):
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;
}
}
如果vector元素的move构造函数可以抛出(这是不太可能的，但仍然是)，那么具有强保证的类似append_move通常不能实现。
