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I have this problem, there is a function foo() as follows,

vector<ClassA> vec;

void foo()
{
    ClassA a;   //inside foo, a ClassA object will be created
    a._ptr = new char[10];

    vec.push_back(a);   //and this newly created ClassA object should be put into vec for later use
}

And AFAIK, vec will invoke ClassA's copy-ctor to make a copy of the newly created object a, and here is the problem. If I define ClassA's copy-ctor the usual way,

ClassA::ClassA(const ClassA &ra) : _ptr(0)
{
    _ptr = ra._ptr;
}

then object a and its copy (created by vec) will have pointers _ptr pointing to the same area, when foo finishes, a will call the destructor to release _ptr, then a's copy in vec will be a dangling pointer, right? Due to this problem, I want to implement ClassA's copy-ctor this way,

ClassA::ClassA(ClassA &ra) : _ptr(0) //take non-const reference as parameter
{
    std::swap(_ptr, a._ptr);
}

Is my implementation ok? Or any other way can help accomplish the job?

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1 Answer

To answer your titular question: Yes, any constructor for a class T that has one mandatory argument of type T & or T const & (it may also have further, defaulted arguments) is a copy constructor. In C++11, there's also a move constructor which requires one argument of type T &&.

Having a non-constant copy constructor that actually mutates the argument gives your class very unusual semantics (usually "transfer semantics") and should be extensively documented; it also prevents you from copying something constant (obviously). The old std::auto_ptr<T> does exactly that.

If at all possible, the new C++11-style mutable rvalue references and move constructors provide a far better solution for the problem of "moving" resources around when they're no longer needed in the original object. This is because an rvalue reference is a reference to a mutable object, but it can only bind to "safe" expressions such as temporaries or things that you have explicitly cast (via std::move) and thus marked as disposable.


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