View C ++ from the perspective of assembly (class inheritance)

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Inheritance is a basic attribute of a class, but how is the function initialized during the class inheritance process? How can we analyze the structure? Let's take a look at the following code?

class employee{public:employee() { printf("employee()!\n");}~employee() { printf("~employee()!\n");}};class manager : public employee{public:manager() { printf("manager()!\n");}~manager() {  printf("~maneger()!\n");}};

Seeing the code above, I believe everyone understands that we have defined such a class. The base class is empoyee, and the inheritance class is Manager. We can see that manager is a special kind of employee. How can we arrange functions during memory construction and analysis?

74:       manager m;00401268   lea         ecx,[ebp-4]0040126B   call        @ILT+60(manager::manager) (00401041)75:   }00401270   lea         ecx,[ebp-4]00401273   call        @ILT+0(manager::~manager) (00401005)00401278   pop         edi00401279   pop         esi0040127A   pop         ebx0040127B   add         esp,44h0040127E   cmp         ebp,esp00401280   call        __chkesp (00408760)00401285   mov         esp,ebp00401287   pop         ebp

We found that the manager structure and analysis structure are also simple. A constructor is constructed when a variable appears. When will the structure be analyzed? The function is analyzed at the end of the process. Next, we can further discuss how the structure and structure of the Manager are actually implemented?

65:   class manager : public employee66:   {67:   public:68:       manager() { printf("manager()!\n");}004012A0   push        ebp004012A1   mov         ebp,esp004012A3   sub         esp,44h004012A6   push        ebx004012A7   push        esi004012A8   push        edi004012A9   push        ecx004012AA   lea         edi,[ebp-44h]004012AD   mov         ecx,11h004012B2   mov         eax,0CCCCCCCCh004012B7   rep stos    dword ptr [edi]004012B9   pop         ecx004012BA   mov         dword ptr [ebp-4],ecx004012BD   mov         ecx,dword ptr [ebp-4]004012C0   call        @ILT+40(employee::employee) (0040102d)004012C5   push        offset string "manager()!\n" (00431020)004012CA   call        printf (004086e0)004012CF   add         esp,4004012D2   mov         eax,dword ptr [ebp-4]004012D5   pop         edi004012D6   pop         esi004012D7   pop         ebx004012D8   add         esp,44h004012DB   cmp         ebp,esp004012DD   call        __chkesp (00408760)004012E2   mov         esp,ebp004012E4   pop         ebp004012E5   ret

We found that the default constructor of employee is added to the Manager constructor. What about the Destructor?

69:       ~manager() {  printf("~maneger()!\n");}00401350   push        ebp00401351   mov         ebp,esp00401353   sub         esp,44h00401356   push        ebx00401357   push        esi00401358   push        edi00401359   push        ecx0040135A   lea         edi,[ebp-44h]0040135D   mov         ecx,11h00401362   mov         eax,0CCCCCCCCh00401367   rep stos    dword ptr [edi]00401369   pop         ecx0040136A   mov         dword ptr [ebp-4],ecx0040136D   push        offset string "~maneger()!\n" (00431040)00401372   call        printf (004086e0)00401377   add         esp,40040137A   mov         ecx,dword ptr [ebp-4]0040137D   call        @ILT+5(employee::~employee) (0040100a)00401382   pop         edi00401383   pop         esi00401384   pop         ebx00401385   add         esp,44h00401388   cmp         ebp,esp0040138A   call        __chkesp (00408760)0040138F   mov         esp,ebp00401391   pop         ebp00401392   ret

We found that when the manager is constructed, the employee is constructed first, and then the manager is printed. When the structure is analyzed, the manager first analyzes itself and then calls the destructor of the employee, the assembly code above demonstrates everything.

[Notice: The following describes virtual functions of the class]