Class member function pointer and mem_fun adapter usage

Let's take a look at the simplest function:

void foo(int a){    cout << a << endl;}

Its function pointer type is

void (*)(int);

We can use this method as follows:

void (*pFunc)(int) = &foo;pFunc(123);

This is the basic usage of function pointers.

 

Class member functions

 

So what are the differences between function pointers for class member functions?

Observe the following classes:

class Foo{public:    //void (Foo::*)(int)    void foo(int a)    {        cout << a << endl;    }    //void (*)(int)    static void bar(int a)    {        cout << a << endl;    }};

We try to use:

void (*pFunc)(int) = &Foo::foo;

Compilation error:

Error: cannot convert 'void (FOO: *) (INT) 'to 'void (*) (INT)' in Initialization

From the above compilation error, we can know that the foo function pointer type is definitely not the expected void (*) (INT), but void (FOO: *) (INT ).

The reason is very simple. The class member function contains an implicit parameter this, So Foo actually has two parameters: Foo * and Int.

Then we try to use the void (FOO: *) (INT) type, as follows:

void (Foo::*pFunc2)(int) = &Foo::foo;

So how can we call it? We use the following two methods:

Foo f;(f.*pFunc2)(45678);

And

Foo *pf = &f;(pf->*pFunc2)(7865);

In this case, the usage is correct.

 

So what are the characteristics of the bar function as a static function?

void (*pFunc)(int) = &Foo::bar; pFunc(123);

We found that the pointer types of static and free functions are the same.

 

Since Foo contains an implicit parameter, can it be converted? We use mem_fun in STL, which is a function adapter.

Foo f;//void (Foo::*)(int) -> void (*)(Foo*, int)(mem_fun(&Foo::foo))(&f, 123);

We can see that mem_func serves as a conversion function, converting the void (FOO: *) (INT) type member function pointer to void (*) (FOO *, INT ), the latter is a pointer of the free function type and can be freely called.

 

Complete.

Class member function pointer and mem_fun adapter usage