二級考試C++基礎(chǔ):return語句的理解

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如果返回的是一個struct對象,return 語句會如何做呢?下面是測試代碼
    #include
    using namespace std;
    struct Big
    {
    char buf[100];
    int i;
    long d;
    }B,B2;
    Big bigfun(Big b)
    {
    b.i=100;
    return b;
    }
    int main()
    {
    B2=bigfun(B);
    return 0;
    }
    在main開頭和結(jié)尾設(shè)斷點
    8: int main()
    19: {
    004012A0 push ebp
    004012A1 mov ebp,esp
    004012A3 sub esp,118h
    //一開始對這迷惑不解,分析了好久
    //原來(118h-40h)剩下的內(nèi)存塊存放了兩個Big的變量
    //低地址放Bigfun()函數(shù)返回值的臨時變量
    //高地址放B2.operator=(Big &) 的參數(shù)
    004012A9 push ebx
    004012AA push esi
    004012AB push edi
    004012AC lea edi,[ebp-118h]
    004012B2 mov ecx,46h
    004012B7 mov eax,0CCCCCCCCh
    004012BC rep stos dword ptr [edi]
    20: B2=bigfun(B);
    004012BE sub esp,6Ch//b參數(shù)入棧
    004012C1 mov ecx,1Bh
    004012C6 mov esi,offset B (00438490)//B的首地址
    004012CB mov edi,esp
    004012CD rep movs dword ptr [edi],dword ptr [esi]//上面這些指令完成對b的初始化工作
    004012CF lea eax,[ebp-0D8h]//eax存放Big類型一個返回值臨時變量的首地址
    004012D5 push eax//注意一般棧調(diào)用沒有這條指令
    004012D6 call @ILT+0(bigfun) (00401005)
    /*
    004012DB add esp,70h
    004012DE mov esi,eax
    004012E0 mov ecx,1Bh
    004012E5 lea edi,[ebp-6Ch]
    004012E8 rep movs dword ptr [edi],dword ptr [esi]
    004012EA mov ecx,1Bh
    004012EF lea esi,[ebp-6Ch]
    004012F2 mov edi,offset B2 (00438500)
    004012F7 rep movs dword ptr [edi],dword ptr [esi]
    21:
    22: return 0;
    004012F9 xor eax,eax
    23:
    24: }
    004012FB pop edi
    004012FC pop esi
    004012FD pop ebx
    004012FE add esp,118h
    00401304 cmp ebp,esp
    00401306 call __chkesp (004081e0)
    0040130B mov esp,ebp
    0040130D pop ebp
    0040130E ret
    */
    @ILT+0(?bigfun@@YA?AUBig@@U1@@Z):
    00401005 jmp bigfun (00401250)
    11: Big bigfun(Big b)
    12: {
    00401250 push ebp
    00401251 mov ebp,esp
    00401253 sub esp,40h
    00401256 push ebx
    00401257 push esi
    00401258 push edi
    00401259 lea edi,[ebp-40h]
    0040125C mov ecx,10h
    00401261 mov eax,0CCCCCCCCh
    00401266 rep stos dword ptr [edi]
    13: b.i=100;
    00401268 mov dword ptr [ebp+70h],64h
    /*
    0012FE50 00 00 00 00 64 00 00 00 00 00 00 00
    */
    14: return b;
    0040126F mov ecx,1Bh
    00401274 lea esi,[ebp+0Ch]//esi=b的首地址
    00401277 mov edi,dword ptr [ebp+8]//edi=返回值臨時變量的首地址
    0040127A rep movs dword ptr [edi],dword ptr [esi]
    0040127C mov eax,dword ptr [ebp+8]
    15: }
    0040127F pop edi
    00401280 pop esi
    00401281 pop ebx
    00401282 mov esp,ebp
    00401284 pop ebp
    00401285 ret
    返回到main函數(shù)
    004012DB add esp,70h//銷毀局部參數(shù)
    004012DE mov esi,eax//bigfunction()返會值臨時變量的首地址
    004012E0 mov ecx,1Bh
    004012E5 lea edi,[ebp-6Ch]//B2.operator=(Big &)參數(shù)的首地址
    004012E8 rep movs dword ptr [edi],dword ptr [esi]
    004012EA mov ecx,1Bh
    004012EF lea esi,[ebp-6Ch]
    004012F2 mov edi,offset B2 (00438500)//B2的首地址
    004012F7 rep movs dword ptr [edi],dword ptr [esi]
    21:
    22: return 0;
    004012F9 xor eax,eax
    23:
    24: }
    004012FB pop edi
    004012FC pop esi
    004012FD pop ebx
    004012FE add esp,118h
    00401304 cmp ebp,esp
    00401306 call __chkesp (004081e0)
    0040130B mov esp,ebp
    0040130D pop ebp
    0040130E ret
    總結(jié):
    1.一般function frame的結(jié)構(gòu)是
    函數(shù)內(nèi)局部變量
    ebp
    eip
    函數(shù)參數(shù)
    如果返回值是一個struct對象function frame
    函數(shù)內(nèi)局部變量
    ebp
    eip
    返回值臨時變量的首地址//特別注意這個
    函數(shù)參數(shù)
    (由于寄存器太小,不可能放一塊struct內(nèi)存,所以保存了臨時變量的首地址).
    2.struct對象的臨時變量是在棧上的,而不是在堆上.