How to store a local copy inside a pointer?

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If i do the following:

C++

void addObject(myObject o, const int& i) {
    o.setI(i);

    this->myVector.push_back(&o);
}

I guess o gets destroyed at the end of the function.

Thus how, can I safely keep a pointer to it in myVector?
Note that, there's no arguing about myVector storing pointers and not objects.

Posted 26-May-14 3:24am

Sergejack

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Stefan_Lang · Accepted Answer · 2014-05-26T04:53:00

Using this definition, o is a temporary variable holding a copy of the object passed to the function, and that temporary will indeed be destroyed upon leaving the function.

There are two possible solutions, depending on what you want:

1. If you're ok to just keep a reference to an existing object, all you need to do is pass the object by reference, rather than by value:

C++

void addObject(myObject& o, const int& i) {

In this case, o is a reference, and &o is a pointer to the object passed to the function.

2. If you want a copy of the original object that survives the scope of the function, you need to create the copy on the heap, rather than the stack. Note that in this case you also should pass the original object by reference - just make it a const reference this time:

C++

void addObject(const myObject& o, const int& i) {
   myObject* po = new myObject(o); // call copy constructor
   po->set(i);
   this->myVector.push_back(po);
}