C ++ const member functions

C ++ const member functions
First fact:

In a certain type, two functions can be declared and defined in this way, and can be overloaded)

void pa(){    cout<<"a"<<endl;}void pa() const{    cout<<"b"<<endl;}

The above statement is correct.

Based on this fact, I thought about its mechanism.

Experiments show that,

The second fact:

A common function (not a member function of a class) can be overloaded as follows ):

void fun(const int &a){    cout<<a<<endl;}void fun(int &a){    cout<<a<<endl;}

The second fact shows that if the function parameter is of the [Reference Type], [const] and [without const] can be reloaded.

Third fact:

A common function (not a member function of a class) can also be overloaded using a pointer like this ):

void test(const int* a){    cout<<*a<<endl;}void test(int* a){    cout<<*a<<endl;}

The third fact shows that the pointer can also be reloaded by adding const and without const.

Analyze the first fact:

According to c ++ primer, a member function itself has an implicit parameter: ClassName * const this // this parameter is irrelevant to the topic of this article.

But let's explain this const first. That is to say, a pointer itself can no longer point to others but to itself.

Therefore, you cannot change the value of this in the class,

This = 0; // error!

Let's talk about these two class member functions:

void pa();void pa() const;

It can be fully written and become a pseudo-code that is easy to understand, like this:

void pa(ClassName *const this);void pa(const ClassName *const this);

First, I have already said this.

The second is a special syntax provided by c ++ to the user, that is, the member function adds const, which is actually added to the front of the implicit parameter this.

Okay. Summary:

This is why member functions can be overloaded by adding the const modifier.