c++ - 如何将方法的引用作为参数传递

Question

如何将方法的引用作为参数传递？它可能看起来像这样:

class Test
{
public:
    Test();

    void Bark();
    void Bark2();

    void TakesABark( void( &method )() );
}

// Start of procedure
Test::Test()
{
    this->TakesABark(Test::Bark);        
    this->TakesABark(Test::Bark2);
}

void Test::Bark()
{
}

void Test::Bark2()
{
}

// Receives a variety of references to methods
void Test::TakesABark( void( &method )() )
{
    // Calls a third party api that would look like this:
    // Barbera::DoThatThingILike(Test::Bark2);
}

Answer 1

class Test
{
public:
    Test();

    void Bark();
    void TakesABark(void (Test::*method)());
};


Test::Test()
{
    this->TakesABark(&Test::Bark);
}