Why Does `push_back` Trigger Multiple Copy Constructor Calls in a Vector?

Vector Push_back and the Mystery of Multiple Copy Constructor Invocations

When working with vectors, one may expect the push_back operation to only necessitate a handful of copy constructor calls. However, certain cases can unveil unexpected behavior, such as multiple calls.

Consider the following code snippet:

<code class="cpp">class Myint {
    int my_int;
public:
    Myint() : my_int(0) { /* ... */ }
    Myint(const Myint& x) : my_int(x.my_int) { /* ... */ }
};

int main() {
    vector<Myint> myints;
    Myint x;

    myints.push_back(x); // Call 1
    x.set(1);
    myints.push_back(x); // Call 2
}</code>

Intuitively, we would expect two copy constructor calls: one for the first push_back and another for the second. However, the output tells a different story:

Call 1 (default constructor)
Call 2 (copy constructor: my_int = 0)
Call 3 (copy constructor: my_int = 0)
Call 4 (copy constructor: my_int = 1)

Why the extra calls?

Internal Vector Dynamics

Internally, a vector typically allocates a certain amount of memory for storage. When the allotted space is exhausted, the vector must reallocate more memory. This reallocation triggers a copy of the existing elements into the new memory.

In the given code, the second push_back necessitates reallocation. Despite the fact that the constructor for Myint is invoked only twice, the copy constructor is utilized three times. This is because the copy constructor is employed to move the first element into the newly allocated memory, initialized with its original value (my_int = 0), and then the copy constructor is invoked again to create the second element, initialized with the updated value (my_int = 1).

Optimizing for Performance

To mitigate these unnecessary copies, it's recommended to:

• Reserve Memory: By increasing the vector's initial capacity, you can prevent reallocation and minimize copy operations.
• Use Emplace_back: This method allows you to directly construct an element in-place, avoiding copies or moves.

In this case, using reserve to preallocate memory would effectively reduce the copy constructor calls to two. Additionally, emplace_back(0) would eliminate the copy altogether, resulting in the optimal performance.

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