What is the order in which the destructors and the constructors are called in C++

Question

What is the order in which the destructors and the constructors are called in C++? Using the examples of some Base classes and Derived Classes

Answer 1

This is clearly described at order-dtors-for-members. Basically, the rule is "First constructed, last destructed."

Constructors calling order:

1. base's constructors are called in the order of appearance after the ":"
2. derived class member's constructors are called in the order of appearance and before class's constructor

Destructors are called in reversed order of called constructors.

Example:

#include 

struct base0 {  base0(){printf("%s\n", __func__);};~base0(){printf("%s\n", __func__);}; };
struct base1 { base1(){printf("%s\n", __func__);}; ~base1(){printf("%s\n", __func__);};};
struct member0 { member0(){printf("%s\n", __func__);};  ~member0(){printf("%s\n", __func__);};};
struct member1 { member1(){printf("%s\n", __func__);}; ~member1(){printf("%s\n", __func__);};};
struct local0 { local0(){printf("%s\n", __func__);}; ~local0(){printf("%s\n", __func__);}; };
struct local1 { local1(){printf("%s\n", __func__);};  ~local1(){printf("%s\n", __func__);};};
struct derived: base0, base1
{
  member0 m0_;
  member1 m1_;
  derived()
  {
    printf("%s\n", __func__);
    local0 l0;
    local1 l1;
  }
  ~derived(){printf("%s\n", __func__);};
};
int main()
{
  derived d;
}

Output:

base0
base1
member0
member1
derived
local0
local1
~local1
~local0
~derived
~member1
~member0
~base1
~base0