Объясните работу кода с перегрузкой операторов

#include <iostream>
 
using namespace std;
 
class A
{
public:
  int number;
  A() : number(0){}
  friend ostream& operator<<(ostream& os, A& a){
    os<<a.number;
  }
  A& operator++(){
    number++;
    return *this;
  }
  A operator++(int){
    A a(*this);
    ++(*this);
    return a;
  }
};
 
int main()
{
  A a1;
 
  cout<<++a1<<", "<<++a1<<endl;
  
  return 0;
}

 cout<<++a1<<", "<<++a1<<endl;

 cout<<a1++<<", "<<++a1<<endl;

 no match for ‘operator<<’ (operand types are ‘std::ostream {aka std::basic_ostream<char>}’ and ‘A’)
   cout<<a1++<<", "<<++a1<<endl;

 friend ostream& operator<<(ostream& os, A& a){
    os<<a.number;
    return os;
  }

#include <iostream>
 
using namespace std;
 
class A
{
public:
  int number;
  A() : number(0){}
  friend ostream& operator<<(ostream& os, A& a){
    std::cout << "operator <<" << std::endl;
    os<<a.number;
  }
  A& operator++(){
    std::cout << "operator ++()" << std::endl;
    number++;
    return *this;
  }
  A operator++(int){
    std::cout << "operator ++(int)" << std::endl;
    A a(*this);
    ++(*this);
    return a;
  }
};
 
int main()
{
 
  { 
    A a1;
    cout<<++a1<<", "<<++a1<<endl;
  } 
  std::cout << "#####################" << std::endl;
  { 
    A a1;
    cout<<++a1;
    cout<<", "<<++a1<<endl;
    std::cout << "#####################" << std::endl;
  } 
 
  return 0;
}

A operator++(int){
    A a(*this);
    ++(*this);
    return a;
  }

A a2 = a++;

const A& operator++(int){
    A a(*this);
    ++(*this);
    return a;
  }