如果类成员是向量，我们是否应该显式地编写一个复制构造函数?

Here ls is holding pointers to char. If a user-defined copy constructor for myType is not provided, will myType's default copy constructor do a deep copy of ls?推荐答案 这里ls持有char的指针.如果未提供复制构造函数，默认的复制构造函数会进行深度复制吗? Here ls is holding pointer to char. If copy constructor is not provided, will default copy constructor do the deep copy?默认的复制构造函数将复制所有成员，即调用其各自的复制构造函数. 1 是的，一个 std :: vector (就C ++而言，没有什么特别的有关)将被正确复制.The default copy constructor will copy all members – i.e. call their respective copy constructors.1 So yes, a std::vector (being nothing special as far as C++ is concerned) will be duly copied.但是，向量中 char * 元素所指向的内存当然不会，因为C ++不知道也不在乎指针指向什么.However, the memory pointed to by the char* elements inside the vector will of course not, since C++ doesn’t know and doesn’t care what the pointers point to.但是这里的解决方案不是提供自定义副本构造函数.它使用数据结构代替原始指针( char * ).碰巧是 std :: string (或根据意图，是 std :: vector< char> ).But the solution here isn’t to provide a custom copy constructor. It’s to use a data structure instead of raw pointers (char*) which does. And this happens to be std::string (or std::vector<char> depending on the intent). 1 这样就创建了复制操作的传递闭包 –这通常是实现深度复制的方式，但是复制操作的实现者总是可以突破的1 Thus creating a transitive closure of the copy operation – this is how deep copying is generally implemented, but of course an implementor of the copy operation can always break out of it. 这篇关于如果类成员是向量，我们是否应该显式地编写一个复制构造函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！