c++的Member Dereferencing Operators功能

感觉这个功能还蛮有意思的，就是不知道什么情况下会用到

#include
using namespace std;

class Example
{
private:
	int feet;
	int inches;
public:
	Example();
	Example(int ft);
	~Example() {};

	void show_in() const;
	void show_ft() const;
	void use_ptr() const;
};

Example::Example()
{
	this->feet = 0;
	this->inches = 0;
}

Example::Example(int ft)
{
	this->feet = ft;
	this->inches = 12 * ft;
}

void Example::show_in() const
{
	cout << this->inches << " inches\n";
}

void Example::show_ft() const
{
	cout << this->feet << " feet\n";
}

void Example::use_ptr() const
{
	Example yard(3);
	int Example::* pt;
	pt = &Example::inches;
	cout << "set pt to &Example::inches:\n";
	cout << "this->*pt: " << this->*pt << endl;
	
	cout << "yard.*pt: " << yard.*pt << endl;

	pt = &Example::feet;
	cout << "set pt to &Example::feet:\n";
	cout << "this->*pt: " << this->*pt << endl;
	cout << "yard.*pt: " << yard.*pt << endl;

	void (Example:: * pf)() const;
	pf = &Example::show_in;
	cout << "set pf to &Example::show_in:\n";
	cout << "using (this->*pf)(): ";
	(this->*pf)();

	cout << "using (yard.*pf)(): ";
	(yard.*pf)();
}

int main()
{
	Example car(15);
	Example van(20);
	Example garage;

	cout << "car.use_ptr() output:\n";
	car.use_ptr();

	cout << "\nvan.use_ptr() output:\n";
	van.use_ptr();

	return 0;
}
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结果

car.use_ptr() output:
set pt to &Example::inches:
this->*pt: 180
yard.*pt: 36
set pt to &Example::feet:
this->*pt: 15
yard.*pt: 3
set pf to &Example::show_in:
using (this->*pf)(): 180 inches
using (yard.*pf)(): 36 inches

van.use_ptr() output:
set pt to &Example::inches:
this->*pt: 240
yard.*pt: 36
set pt to &Example::feet:
this->*pt: 20
yard.*pt: 3
set pf to &Example::show_in:
using (this->*pf)(): 240 inches
using (yard.*pf)(): 36 inches
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