std::vector memory handling

Question

I tried to google and search an answer to my question, but I couldn't find any valid explanation hence I am posting my question here. Following is my sample code and output:

#include <iostream> #include "vector" using namespace std;  typedef struct Node{     int data;     Node(){         data = 0;         std::cout << "Node created. " << this <<'\n';     }     ~Node(){         data = 0;         std::cout << "Node destroyed. " << this <<'\n';     } } Node;  int main() {     std::vector<Node> vec;     for(int i = 0; i < 2 ; i++)        vec.push_back( *(new Node));     return 0; } 

Output:

Node created. 0x9e0da10 Node created. 0x9e0da30 Node destroyed. 0x9e0da20 Node destroyed. 0x9e0da40 Node destroyed. 0x9e0da44 

Why is there an extra destroy and why are created objects different from destroyed object?

Answer 1

vec.push_back( *(new Node)); is an immediate memory leak.

First you dynamically allocate a Node, then you copy that Node into the vector. The copy operation is what creates the new object, which is why this is different.

The original (dynamically allocated) Node is never deallocated, but the copies are when the destructor of the vector runs (i.e. at the end of the function).

Why three destructor calls instead of two? That's caused by the automatic reallocation of the vector when you push_back. It moves/copies its elements to a new memory location, destroying the old elements.

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Note that normally, when you simply want a vector of n default-constructed elements, you'd do:

std::vector<Node> vec(2); 

This calls Node() (the default constructor) for elements vec[0] and vec[1], and you don't need a loop (or dynamic allocation).