C++ capture value in lambda

Question

I have the following example to capture a variable in lambda. If I capture it by value, the printed value is 12 which is expected. If I capture it by reference, the printed value is 02. I don't understand why. I wanted to use capture by reference to avoid copy.

class test {
    int i_;
  public:
    test(int i) : i_(i) {}

    void see() const {
        std::cout << i_ << std::endl;
    } 
};

int main ()
{
  std::vector<test> vec;
  vec.emplace_back(1);
  // expected 
  // auto f1 = [op{vec.back()} () { 
  //     op.see()
  // }

  // Not expected
  auto f1 = [op{&vec.back()}]() {
      op->see();
  };
  
  vec.emplace_back(2);
  auto f2 = [op{&vec.back()}]() {
      op->see();
  };
  
  f1();
  f2();
  return 0;
}

score 5 · Answer 1 · answered Sep 16 '21 at 18:55

5

emplace_back() potentially reallocates, and reallocation of a std::vector invalidates all existing iterators, pointers, and references, including the captured reference.

answered Sep 16 '21 at 18:55

Ben Voigt

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This also applies to push_back, for anyone curious. – Andrew Holmgren Sep 16 '21 at 19:03
3

Additional reading: [Iterator invalidation rules for C++ containers](https://stackoverflow.com/questions/6438086/iterator-invalidation-rules-for-c-containers) – user4581301 Sep 16 '21 at 19:06
1

@Andrew: Also `insert()`, `push_front()` and a bunch of others. – Ben Voigt Sep 16 '21 at 22:03

C++ capture value in lambda

1 Answers1