How do you move unique_ptr out of the vector <unique_ptr <Foo>>?

I would like to move unique_ptr<Foo>

from vector<unique_ptr<Foo>>

. Consider my code:

#include <vector>
#include <memory>
#include <iostream>

using namespace std;

class Foo {
public:
  int x;
  Foo(int x): x(x) {};
  ~Foo() {
    cout << "Destroy of id: " << x << "\n";
    x = -1;
  };
};

int main(int argc, char *argv[]) {
  auto foos = vector<unique_ptr<Foo>>();
  foos.push_back(unique_ptr<Foo>(new Foo(100)));
  foos.push_back(unique_ptr<Foo>(new Foo(101)));
  foos.push_back(unique_ptr<Foo>(new Foo(102)));

  // Print all
  cout << "Vector size: " << foos.size() << "\n";
  for (auto i = foos.begin(); i != foos.end(); ++i) {
    cout << (*i)->x << "\n";
  }

  // Move Foo(100) out of the vector
  {
    auto local = move(foos.at(0));
    cout << "Removed element: " << local->x << "\n";
  }

  // Print all! Fine right?
  cout << "Vector size: " << foos.size() << "\n";
  for (auto i = foos.begin(); i != foos.end(); ++i) {
    cout << (*i)->x << "\n";
  }

  return 0;
}

      

I expected it to give:

Vector size: 3
100
101
102
Removed element: 100
Destroy of id: 100
Vector size: 2
101
102

      

But instead I get this result:

Vector size: 3
100
101
102
Removed element: 100
Destroy of id: 100
Vector size: 3
Segmentation fault: 11

      

Why is my vector size 3 and why am I getting segmentation error? How can I get the desired result?

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1 answer


Simplify your question to:

vector<unique_ptr<Foo>> foos;
foos.push_back(unique_ptr<Foo>(new Foo(100)));
auto local = std::move(foos[0]);
std::cout << foos[0]->x << '\n';

      

After creating local

, moving foos[0]

, foos[0]

no longer has the ownership of the pointer. It is empty. The expression is dereferenced, this becomes undefined behavior, which in your case manifests itself as a segmentation fault. At this point it is vector

perfectly "intact", it contains one empty unique_ptr

and is equivalent in state:

vector<unique_ptr<Foo>> foos(1);

      



You should just check that your unique_ptr

owns the pointer before dereferencing it:

if (foos[0]) {
    // we wouldn't get here
    std::cout << foos[0]->x << '\n';
}

      

Alternatively, since you want to apply the invariant to what yours vector

contains only valid pointers, only erase

this element should be as part of a branch operation from it :

auto local = std::move(foos[0]);
foos.erase(foos.begin());
// now foos is empty

      

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