Consider the following code example:
vector<int> v; // Add some elements for(int i=0; i<10; ++i) v.push_back(i); int* my_favorite_element_ptr = &v; cout << "My favorite element = " << (*my_favorite_element_ptr) << endl; cout << "Its address = " << my_favorite_element_ptr << endl; cout << "Adding more elements…"<< endl; // Adding more elements for(int i=0; i<100; ++i) v.push_back(i*10); cout << "My favorite element = " << (*my_favorite_element_ptr) << endl; cout << "Its address = " << &v << endl;
What’s going on here? We create a vector containing 10 elements, and for some reason decide to save for later a pointer to element with index 3. Then we add more elements to the vector and try to reuse the pointer we’ve acquired before. What is wrong with this code? Let’s look at the output it produces:
My favorite element = 3 Its address = 0x1001000cc Adding more elements … My favorite element = 3 Its address = 0x10010028c
Note that after we add more elements to the vector, the address of the
&v has changed! The
problem is that when we add new elements to the vector, the existing
elements might move to a totally different location.
Here is how such code works. When we create a vector, it allocates by default some number of elements (usually about 16). Then if we try to add more elements than the capacity allows, the vector allocates a new, larger array, copies existing elements from the old location to a new one, and continues ...