0

在ROS中有一个函数叫做NodeHanle::subscribe(Args...)：NodeHandle::subscribe。这使您可以将 PRIVATE 成员函数作为回调传递。

但是，当我自己尝试时（使用 std::bind 传递私有成员函数），我的编译器总是失败并抱怨Foo::foo() is a private member function. 当我Foo::foo转为公共职能时，一切都恢复正常。

template<typename T>
void getWrapper1(void(T::*fn)(int), T *t) { 
  return [&](int arg) {
    std::cout << "process before function with wrapper" << std::endl;
    (t->*fn)(arg);
    std::cout << "process after function with wrapper" << std::endl;
  };
}

void getWrapper2(std::function<void(int)> fn) {
  return [=](int arg) {
    std::cout << "process before function with wrapper" << std::endl;
    fn(arg);
    std::cout << "process after function with wrapper" << std::endl;
  }
}

class Foo {
private:
  void foo(int a) {
    std::cout << __FUNCTION__ << a << std::endl;
  }
}

int main(int argc, char** argv) {
  Foo foo_inst;
  auto func1 = getWrapper1(&Foo::foo, &foo_inst); // fail because foo is private
  auto func2 = getWrapper2(std::bind(&Foo::foo, &foo_inst, std::placeholders::_1));  // fail because foo is private
  func1(1);
  func2(2);
  return 0;
}

从这个答案，使用std::function也可以传递私有成员函数。但是我尝试过的不同。

值得一提的是，在getWrapper2我使用[=]而不是[&]因为使用[&]可能会导致段错误。为什么它必须是“价值捕获”？

平台：GCC 5.4.0、c++14、ubuntu16.04

4

1 回答 1

3

You must pass it from the inside. You cannot access private function from the outside of the class. Not even pointer to private stuff. Private is private.

class Foo {
    void foo(int a) {
        std::cout << __FUNCTION__ << a << std::endl;
    }

 public:
    auto getWrapper() {
        // using a lambda (recommended)
        return getWrapper2([this](int a) {
            return foo(a);
        });

        // using a bind (less recommended)
        return getWrapper2(std::bind(&Foo::foo, this, std::placeholders::_1));
    }
}

Why it has to be a "value capture"?

Both wrapper need to value capture. Your Wrapper1 have undefined behaviour.

Consider this:

// returns a reference to int
auto test(int a) -> int& {
    // we return the local variable 'a'
    return a;
    // a dies when returning
}

The same thing happen with a lambda:

auto test(int a) {
    // we capture the local variable 'a'
    return [&a]{};
    // a dies when returning
}

auto l = test(1);
// l contain a captured reference to 'a', which is dead

Pointers are passed by value. A pointer is itself an object. A pointer has itself a lifetime and can die.

auto test(int* a) -> int*& {
    // we are still returning a reference to local variable 'a'.
    return a;
}

And... you guessed it, the same thing for std::function:

auto test(std::function<void(int)> a) {
    // return a lambda capturing a reference to local variable 'a'.
    return [&a]{};
}
于 2019-08-22T18:18:47.847 回答