用汇编的眼光看C++（之类继承） - c++编程基础

A7 push esi

004012A8 push edi

004012A9 push ecx

004012AA lea edi,[ebp-44h]

004012AD mov ecx,11h

004012B2 mov eax,0CCCCCCCCh

004012B7 rep stos dword ptr [edi]

004012B9 pop ecx

004012BA mov dword ptr [ebp-4],ecx

004012BD 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,4

004012D2 mov eax,dword ptr [ebp-4]

004012D5 pop edi

004012D6 pop esi

004012D7 pop ebx

004012D8 add esp,44h

004012DB cmp ebp,esp

004012DD call __chkesp (00408760)

004012E2 mov esp,ebp

004012E4 pop ebp

004012E5 ret

我们发现，manager的构造里面添加了employee的缺省构造函数，那么析构函数呢？

copy to clipboardprint 69: ~manager() { printf("~maneger()!\n");}

00401350 push ebp

00401351 mov ebp,esp

00401353 sub esp,44h

00401356 push ebx

00401357 push esi

00401358 push edi

00401359 push ecx

0040135A lea edi,[ebp-44h]

0040135D mov ecx,11h

00401362 mov eax,0CCCCCCCCh

00401367 rep stos dword ptr [edi]

00401369 pop ecx

0040136A mov dword ptr [ebp-4],ecx

0040136D push offset string "~maneger()!\n" (00431040)

00401372 call printf (004086e0)

00401377 add esp,4

0040137A mov ecx,dword ptr [ebp-4]

0040137D call @ILT+5(employee::~employee) (0040100a)

00401382 pop edi

00401383 pop esi

00401384 pop ebx

00401385 add esp,44h

00401388 cmp ebp,esp

0040138A call __chkesp (00408760)

0040138F mov esp,ebp

00401391 pop ebp

00401392 ret

69: ~manager() { printf("~maneger()!\n");}