[C ++ in-depth analysis learning summary] 24 classic problem analysis two

[C ++ in-depth analysis learning summary] 24 classic problem analysis two

Author CodeAllen , please indicate the source

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1. Questions about destruction
When multiple objects exist in the program, how to determine the destruction order of these objects?

The order in which destructors are called when a single object is created

• 1. Call the destructuring process of the parent class (explained in subsequent courses)
• 2. Call the member variable destructor (the calling order is the same as the declaration order)
• 3. Call the destructor of the class itself
  // the destructor is in the reverse order of the corresponding constructor.

Construction and destruction order

#include <stdio.h>

class Member
{
    const char* ms;
public:
    Member(const char* s)
    {
        printf("Member(const char* s): %s\n", s);
        
        ms = s;
    }
    ~Member()
    {
        printf("~Member(): %s\n", ms);
    }
};

class Test
{
    Member mA;
    Member mB;
public:
    Test() : mB("mB"), mA("mA")
    {
        printf("Test()\n");
    }
    ~Test()
    {
        printf("~Test()\n");
    }
};

Member gA("gA");


int main()
{
    Test t;
    
    return 0;
}

2. Answers about destructuring
For stack objects and global objects, similar to the order of stacking and stacking, the last constructed object is destructed first! !
The destruction of the heap object occurs when delete is used, and it is related to the order in which delete is used! !

3. Questions about const objects
Can the const keyword modify class objects? If so, what are the characteristics?

• const keyword can modify objects
• Const modified objects are read-only objects
• Member variables of read-only objects are not allowed to be changed
• The read-only object is a concept of the compilation stage, and is invalid at runtime

Const member function in C ++

• const objects can only call const member functions
• Only const member functions can be called in const member functions
• const member function can not directly rewrite the value of member variables

Definition of const member function:

• Type ClassName::function(Type p)const
• After the member function declaration, before the function body
• The function declaration in the class and the actual function definition must carry the const keyword

On the const function of class

#include <stdio.h>

class Test
{
    int mi;
public:
    int mj;
    Test(int i);
    Test(const Test& t);
    int getMi();
};

Test::Test(int i)
{
    mi = i;
}

Test::Test(const Test& t)
{
    
}
    
int Test::getMi()
{
    return mi;
}

int main()
{
    const Test t(1);
    
    t.mj = 1000;  //会报错，因为const修饰对象使其变为只读，不能出现在赋值符号左边
                  //
    return 0;
}

General projects do not define so many const functions, so you can directly call t.mi; (this is a private variable, but you can call it directly)

4.
Are member functions and member variables in question about class members belonging to specific objects?
From an object-oriented perspective

• Objects are composed of attributes (member variables) and methods (member functions)
  from the perspective of program operation
• Objects are composed of data and functions
  • Data can be located on the stack, heap and global data area
  • The function can only be located in the code segment

in conclusion

• Each object has its own independent properties (member variables)
• All objects share class methods (member functions)
• Methods can directly access the properties of objects
• The hidden parameter this in the method is used to refer to the current object

Member function secret

#include <stdio.h>

class Test
{
    int mi;
public:
    int mj;
    Test(int i);
    Test(const Test& t);
    int getMi();
    void print();
};


Test::Test(int i)
{
    mi = i;
}


Test::Test(const Test& t)
{
    mi = t.mi;
}
    
int Test::getMi()
{
    return mi;
}


void Test::print()
{
    printf("this = %p\n", this);
}


int main()
{
    Test t1(1);
    Test t2(2);
    Test t3(3);
    
    printf("t1.getMi() = %d\n", t1.getMi());
    printf("&t1 = %p\n", &t1);
    t1.print();
    
    printf("t2.getMi() = %d\n", t2.getMi());
    printf("&t2 = %p\n", &t2);
    t2.print();
    
    printf("t3.getMi() = %d\n", t3.getMi());
    printf("&t3 = %p\n", &t3);
    t3.print();
    
    return 0;
}

/*
t1.getMi() = 1
&t1 = 0x7fffe4bb2810
this = 0x7fffe4bb2810  
t2.getMi() = 2
&t2 = 0x7fffe4bb2818
this = 0x7fffe4bb2818
t3.getMi() = 3
&t3 = 0x7fffe4bb2820
this = 0x7fffe4bb2820
*/

Summary
destructor order opposite to the order of objects configured
const keyword can modify the object, the object to obtain a read-only
read-only objects can call member functions const
member functions of the class of all objects share
hidden this pointer used to indicate the current object