Are C ++ 11 lambdas themselves an RAII auto-release object?

Question

I would like to write a class method that, if necessary, accepts a lambda to customize its behavior. So when I use this class, I wonder if I need to worry that the lambda itself is out of scope?

The lambda will not refer to any external vars, so I don’t worry about the scale of the vars, only about the area of ​​the lambda itself that I would keep in the class.

Do I need to worry about how / where the lambda itself was created?

Answer 1

Links, outside of narrow situations, do not extend the life of the item you are linking to.

, , - undefined.

undefined lambda " this", . , , , .

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std::function<void()> , . , . std::function<void()> , , , .

Answer 2

, , RAII . . -

template <class T> class locker {
private:
  mutable T m_t; // Copies the lambda here.
  mutable std::mutex m_m;
public:
  locker( T t = T{} ) : m_t(t) {}
  template <typename F>
  auto operator()(F f) const -> decltype(f(m_t)) {
    std::lock_guard<mutex> _{m_m};
    return f(t);
  }
};


// usage 
locker<std::string> s;
s([](string &s) {
  s += "foobar";
  s += "barfoo";
});

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