从逆向分析角度看C++拷贝构造函数

　　一段C++代码：

//: HowMany_2.cpp
#include <iostream>

using namespace std;

class HowMany {
  static int objectCount;
 
public:
  HowMany() {
    ++objectCount;
    print("HowMany()");
  }
  ~HowMany() {
    --objectCount;
    print("~HowMany()");
  }
  HowMany(const HowMany& h) {
    ++objectCount;
    print("HowMany(const HowMany&)");
  }
 
  void print(const char ss[]) {
    cout << ss << ": ";
    cout << "objectCount = " << objectCount << endl;
    return ;
  }
};

int HowMany::objectCount = 0;

HowMany f(HowMany x) {
  x.print("x argument inside f()");
  cout << "Return From f()" << endl;
  return x;  // 有返回值 x
}

int main() {
  {
    HowMany h;
    cout << "Entering f()" << endl;
    HowMany h2 = f(h);
  }
  return 0;
} ///:~

　　运行结果：

　　Assembly Code：

38:   int main() {
39:     {
40:       HowMany h;
004017FD   lea         ecx，[h]     ; [h] 为对象 h 的内存地址
00401800   call        @ILT+685(HowMany::HowMany) (004012b2) ; 调用构造函数
00401805   mov         dword ptr [ebp-4]，0
41:       cout << "Entering f()" << endl;
0040180C   push        offset @ILT+200(std::endl) (004010cd)
00401811   push        offset string "Entering f()" (0046f090)
00401816   push        offset std::cout (0047ce98)
0040181B   call        @ILT+660(std::operator<<) (00401299) ; 题外话，观察一下进栈顺序
00401820   add         esp，8
00401823   mov         ecx，eax
00401825   call        @ILT+480(std::basic_ostream<char，std::char_traits<char> >::operator<<) (004011e5)
42:       HowMany h2 = f(h);
0040182A   push        ecx
0040182B   mov         ecx，esp     ; 当前 ESP 所指的栈块作为临时对象Temp的内存地址
0040182D   mov         dword ptr [ebp-18h]，esp
00401830   lea         eax，[h]    
00401833   push        eax     ; 将h的内存地址[h]压入堆栈
00401834   call        @ILT+0(HowMany::HowMany) (00401005) ; 调用拷贝构造函数，把h的内容拷贝到Temp的内存中
00401839   mov         dword ptr [ebp-1Ch]，eax
0040183C   lea         ecx，[h2]
0040183F   push        ecx     ; 将h2的内存地址[h2]压入堆栈
00401840   call        @ILT+640(f) (00401285)   ; 调用f()函数
00401845   add         esp，8
00401848   mov         dword ptr [ebp-20h]，eax
43:     }
0040184B   lea         ecx，[h2]
0040184E   call        @ILT+500(HowMany::~HowMany) (004011f9) ; 调用析构函数，销毁h2
00401853   mov         dword ptr [ebp-4]，0FFFFFFFFh
0040185A   lea         ecx，[h]
0040185D   call        @ILT+500(HowMany::~HowMany) (004011f9) ; 调用析构函数，销毁h
44:     // getchar();
45:     return 0;
00401862   xor         eax，eax
46:   } ///:~