如何使用 std::reference_wrapper<t> ：：操作员（）</t>

Question

I'd like to use the std::reference_wrapper<T>::operator() "the same" as std::reference_wrapper<T>::get , but following example fails for operator()

#include <cstdint>
#include <functional>

class Foo {
public:
  void Print() {
    std::printf("Foo\n");
  }
};

class Bar {
public:
  Bar(Foo &foo): wrapper{foo} {}
  void Print() {
    wrapper.get().Print();  // OK
    // wrapper().Print();   // FAIL
  }
private:
  std::reference_wrapper<Foo> wrapper;
};

int main() {
  Foo foo{};
  Bar bar{foo};
  bar.Print();
  return 0;
}

Is this possible? Where is my misunderstanding?

Thanks for the help

Zlatan

Answer 1

operator() of std::reference_wrapper does not do the same as .get() .

Its operator() invokes a function or other Callable object to which the stored reference refers.

In your example a Foo object is not a Callable . If Print was instead operator() , then you could simply call it with

wrapper();

Answer 2

Is this possible? Where is my misunderstanding?

No. std::reference_wrapper<T>::operator() only exists when T::operator() exists, and it simply calls that, forwarding the arguments provided.

Are you mistaking it for std::reference_wrapper<T>::operator T& ?

class Bar {
public:
  Bar(Foo &foo): wrapper{foo} {}
  void Print() {
    Foo & f = wrapper;
    f.Print()
  }
private:
  std::reference_wrapper<Foo> wrapper;
};