Question

我有以下代码片段

#include <iostream>
#include <vector> 

class Move {
    private:
        int* data;
         
    public: 
        void set_data_value(int d) {
            *data = d; 
        }
        int get_data_value(){
            return *data; 
        }

        Move(int d);
        Move(const Move &source);
        ~Move();
};

Move::Move(int d) {
    std::cout << "ctor being called" << std::endl; 
    data = new int; 
    *data = d; 
}

Move::Move(const Move& source) {
    std::cout << "copy ctor being called" << std::endl;
    data = new int; 
    *data = *source.data;
}

Move::~Move() {
    std::cout << "deleting Move" << std::endl; 
    delete data; 
}
int main() {
    Move x{1};
    
    Move y{x}; 

    std::vector<Move> vec; 
    vec.push_back(x);
    vec.push_back(y);

    return 1; 

}

输出如下：

ctor being called
copy ctor being called
copy ctor being called
copy ctor being called
copy ctor being called
deleting Move
deleting Move
deleting Move
deleting Move
deleting Move

我的问题是：当我只看到 3 个副本时，为什么要调用 4 次复制构造函数？（1 个来自 Move y{x}，2 个来自两个 push_back() 调用）

Answer 1

您错过了计算向量的重新分配。将此行添加到您的代码中：

int main() {
    Move x{1};
    
    Move y{x}; 

    std::vector<Move> vec; 
    std::cout << "capacity: " << vec.capacity() << "\n";
    vec.push_back(x);
    std::cout << "capacity: " << vec.capacity() << "\n";
    vec.push_back(y);
    std::cout << "capacity: " << vec.capacity() << "\n";

    return 1; 

}

那么可能的输出是：

ctor being called
copy ctor being called
capacity: 0
copy ctor being called
capacity: 1
copy ctor being called
copy ctor being called
deleting Move
capacity: 2
deleting Move
deleting Move
deleting Move
deleting Move

vector 的容量从 0 开始，然后根据需要增加（常见的因子为 2 或 3）。可以看到，当容量从 1 增加到 2 时，会调用复制构造函数，将向量中已经存在的元素复制到内存中的其他位置。

注意上面的输出可以不同。分配策略需要进行细微的优化。它要求 push_back 具有摊销不变的复杂性，这意味着容量不能线性增长。虽然因子不一定是2。某些实现使用 3 或其他因素。

Answer 2

这是由于每次数组不再足够时向量增加容量您可以通过保留一些空间来防止这种情况 reserve()

https://en.cppreference.com/w/cpp/container/vector/reserve

为什么复制构造函数被调用两次？

2 个答案: