Unexpected behavior of std::vector::operator= in C++

Question

I was expected that with

v2 = v1

the content of v1 is copied into v2.

What I experience is that the content of v1 is not copied into v2 if the content of v1 was not initialized in constructor.

Just see the code:

#include <iostream>
#include <vector>

using namespace std;

int main (void) {
  vector<int> v1; // if I use vector<int> v1(10,0); it just works fine
  vector<int> v2;
  v1.reserve(5);
  v2.reserve(5);
  for (int i = 0; i < 5; i++) v1[i] = i*i;
  v2 = v1;
  for (int i = 0; i < 5; i++) {
    cout << "v1[" << i << "] = " << v1[i] << endl;
    cout << "v2[" << i << "] = " << v2[i] << endl;
  }
  return 0;
}

The result of the execution is:

v1[0] = 0
v2[0] = 0
v1[1] = 1
v2[1] = 0
v1[2] = 4
v2[2] = 0
v1[3] = 9
v2[3] = 0
v1[4] = 16
v2[4] = 0

Why this strange behaviour?

Answer 1

Your expectation is correct. v2 = v1; copies all the elements of v1 into v2 . The problem is that you are accessing your vectors out of bounds. This is undefined behaviour .

std::vector::reserve does not increase their size. It simply allocates more storage for the size to grow into as needed. You need to call std::vector::resize instead, initialize the vector to have the right size, or call push_back in the loop.

Note that you should use the size() member variable, the begin() and end() iterators, or a range-based for-loop to iterate over a vector. That will ensure you do not access out of bounds.