c++ - 将 unique_ptr 的容器传递给构造函数？

Question

我在这里错过了什么？为什么我不能将 vector 作为类构造函数的一部分 move ？从构造函数中删除 const 也无济于事。

#include <iostream>
#include <vector>
#include <memory>

using namespace std;

class Bar
{
public:
  Bar(const vector<unique_ptr<char>> vec);
  vector<unique_ptr<char>> vec_;
};

Bar::Bar(const vector<unique_ptr<char>> vec) :
  vec_(move(vec)) //not ok
{
}

int main()
{
  vector<unique_ptr<char>> vec;
  vec.push_back(unique_ptr<char>(new char('a')));
  vec.push_back(unique_ptr<char>(new char('b')));
  vec.push_back(unique_ptr<char>(new char('c')));
  vector<unique_ptr<char>> vec1 (move(vec)); //ok
  Bar bar(vec1);
  return 0;
}

Answer 1

以下should compile fine :

#include <iostream>
#include <vector>
#include <memory>

using namespace std;

class Bar
{
public:
  Bar(vector<unique_ptr<char>> vec);
  vector<unique_ptr<char>> vec_;
};

Bar::Bar(vector<unique_ptr<char>> vec) : // If you intend to move something,
                                         // do not make it const, as moving
                                         // from it will in most cases change
                                         // its state (and therefore cannot be
                                         // const-qualified).
  vec_(move(vec))
{
}

int main()
{
  vector<unique_ptr<char>> vec;
  vec.push_back(unique_ptr<char>(new char('a')));
  vec.push_back(unique_ptr<char>(new char('b')));
  vec.push_back(unique_ptr<char>(new char('c')));
  vector<unique_ptr<char>> vec1 (move(vec));
  Bar bar(std::move(vec1)); // Just like the line immediately above,
                            // the explicit `move` is required, otherwise
                            // you are requesting a copy, which is an error.
  return 0;
}

我保留了您的其余代码不变，但您可能需要阅读 Why is “using namespace std;” considered bad practice?