std :: shared_ptr深层复制对象 [英] std::shared_ptr deep copy object

问题描述

在C ++ 11上找不到很多，只能在增强上使用。

Can't find much on that for C++11 but only on boost.

请考虑以下课程：

class State
{
   std::shared_ptr<Graph> _graph;

 public:

    State( const State & state )
    {
        // This is assignment, and thus points to same object
        this->_graph = std::make_shared<Graph>( state._graph ); 

        // Deep copy state._graph to this->_graph ?
        this->_graph = std::shared_ptr<Graph>( new Graph( *( state._graph.get() ) ) );

        // Or use make_shared?
        this->_graph = std::make_shared<Graph>( Graph( *( state._graph.get() ) ) );
    }   
};

假设 class Graph 确实具有副本构造函数：

Suppose class Graph does have a copy constructor:

Graph( const Graph & graph )

我不想让 this-> _ graph 指向/共享同一对象！
相反，我希望 this-> _ graph 将对象从 state._graph 复制到我自己的 this-> _ graph

I do not want to have this->_graph point/share the same object! Instead, I want this->_graph to deep copy the object from state._graph, into my own this->_graph duplicate.

上面的方法是否正确？

std :: make_shared的文档指出：

此外，如果g抛出异常，f（shared_ptr（new int（42）），g（））可能导致内存泄漏
。如果使用了make_shared
，则不存在此问题。

Moreover, f(shared_ptr(new int(42)), g()) can lead to memory leak if g throws an exception. This problem doesn't exist if make_shared is used.

还有另一种解决方法，更安全或更安全

Is there another way of going about this, safer or more reliable?

推荐答案

如果要复制 Graph 复制对象时，始终可以定义副本构造函数和赋值运算符来做到这一点：

If you want to make a copy of the Graph object when you make a copy of the object, you can always define your copy constructor and assignment operator to do just that:

State::State(const State& rhs) : _graph(std::make_shared(*rhs._graph)) {
   // Handled by initializer list
}
State::State(State&& rhs) : _graph(std::move(rhs._graph)) {
   // Handled by initializer list
}
State& State::operator= (State rhs) {
    std::swap(*this, rhs);
    return *this;
}

希望这会有所帮助！

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