Question

我将 std::vector 称作 c 数组的安全封装。在内部实现上，它和 std::string 相似，包含指向缓冲区的指针，指向数组尾部的指针，以及指向缓冲区尾部的指针。 std::vector 中的元素在内存中连续存放，和通常中的数组一样。在 C++11 包含一个新方法.data()，它返回指向缓冲区的指针，这和 std::string 中的.c_str() 相同。 在堆中分配的缓冲区大小将超过数组自身大小。 MSVC 和 GCC 的实现相似，只是结构体中元素名稍有不同，因此下面的代码在两种编译器上都能工作。下面是用于转储 std::vector 结构的类 C 代码。

#include <stdio.h>
#include <vector>
#include <algorithm>
#include <functional>
struct vector_of_ints
{
// MSVC names:
int *Myfirst;
int *Mylast;
int *Myend;
// GCC structure is the same, names are: _M_start, _M_finish, _M_end_of_storage
};
void dump(struct vector_of_ints *in)
{
printf ("_Myfirst=%p, _Mylast=%p, _Myend=%p\n", in->Myfirst, in->Mylast, in->Myend);
size_t size=(in->Mylast-in->Myfirst);
size_t capacity=(in->Myend-in->Myfirst);
printf ("size=%d, capacity=%d\n", size, capacity);
for (size_t i=0; i<size; i++)
printf ("element %d: %d\n", i, in->Myfirst[i]);
};
int main()
{
std::vector<int> c;
dump ((struct vector_of_ints*)(void*)&c);
c.push_back(1);
dump ((struct vector_of_ints*)(void*)&c);
c.push_back(2);
dump ((struct vector_of_ints*)(void*)&c);
c.push_back(3);
dump ((struct vector_of_ints*)(void*)&c);
c.push_back(4);
dump ((struct vector_of_ints*)(void*)&c);
c.reserve (6);
dump ((struct vector_of_ints*)(void*)&c);
c.push_back(5);
dump ((struct vector_of_ints*)(void*)&c);
c.push_back(6);
dump ((struct vector_of_ints*)(void*)&c);
printf ("%d\n", c.at(5)); // bounds checking
printf ("%d\n", c[8]); // operator[], no bounds checking
};
如果编译器是 MSVC，下面是输出样例。
_Myfirst=00000000, _Mylast=00000000, _Myend=00000000
size=0, capacity=0
_Myfirst=0051CF48, _Mylast=0051CF4C, _Myend=0051CF4C
size=1, capacity=1
element 0: 1
_Myfirst=0051CF58, _Mylast=0051CF60, _Myend=0051CF60
size=2, capacity=2
element 0: 1
element 1: 2
_Myfirst=0051C278, _Mylast=0051C284, _Myend=0051C284
size=3, capacity=3
element 0: 1
element 1: 2
element 2: 3
_Myfirst=0051C290, _Mylast=0051C2A0, _Myend=0051C2A0
size=4, capacity=4
element 0: 1
element 1: 2
element 2: 3
element 3: 4
_Myfirst=0051B180, _Mylast=0051B190, _Myend=0051B198
size=4, capacity=6
element 0: 1
element 1: 2
element 2: 3
element 3: 4
_Myfirst=0051B180, _Mylast=0051B194, _Myend=0051B198
size=5, capacity=6
element 0: 1
element 1: 2
element 2: 3
element 3: 4
element 4: 5
_Myfirst=0051B180, _Mylast=0051B198, _Myend=0051B198
size=6, capacity=6
element 0: 1
element 1: 2
element 2: 3
element 3: 4
element 4: 5
element 5: 6
6
6619158

可以看到，在 main 函数的头部也没有分配空间。当第一次 push_back() 调用结束后，缓冲区被分配。同时在每一次调用 push_back() 后，数组的大小和缓冲区容量都增大了。缓冲区地址也变化了，因为 push_back() 函数每次都会在堆中重新分配缓冲区。这是个耗时的操作，这就是为什么提前预测数组大小，同时调用.reserve() 方法预留空间比较重要了。最后数据是垃圾数据，没有数组元素位于这个位置，因此打印出了随机的数据。这也说明了 std::vector 的[]操作并不校验数组下标是否越界。然而.at() 方法会做相应检查，当出现越界时会抛出 std::out_of_range。 让我们来看代码：

Listing 34.11: MSVC 2012 /GS- /Ob1

$SG52650 DB ’%d’, 0aH, 00H
$SG52651 DB ’%d’, 0aH, 00H
_this$ = -4 ; size = 4
__Pos$ = 8 ; size = 4
?at@?$vector@HV?$allocator@H@std@@@std@@QAEAAHI@Z PROC ; std::vector<int,std::allocator<int> >::
at, COMDAT
; _this$ = ecx
push ebp
mov ebp, esp
push ecx
mov DWORD PTR _this$[ebp], ecx
mov eax, DWORD PTR _this$[ebp]
mov ecx, DWORD PTR _this$[ebp]
mov edx, DWORD PTR [eax+4]
sub edx, DWORD PTR [ecx]
sar edx, 2
cmp edx, DWORD PTR __Pos$[ebp]
ja SHORT $LN1@at
push OFFSET ??_C@_0BM@NMJKDPPO@invalid?5vector?$DMT?$DO?5subscript?$AA@
call DWORD PTR __imp_?_Xout_of_range@std@@YAXPBD@Z
$LN1@at:
mov eax, DWORD PTR _this$[ebp]
mov ecx, DWORD PTR [eax]
mov edx, DWORD PTR __Pos$[ebp]
lea eax, DWORD PTR [ecx+edx*4]
$LN3@at:
mov esp, ebp
pop ebp
ret 4
?at@?$vector@HV?$allocator@H@std@@@std@@QAEAAHI@Z ENDP ; std::vector<int,std::allocator<int> >::
at
_c$ = -36 ; size = 12
$T1 = -24 ; size = 4
$T2 = -20 ; size = 4
$T3 = -16 ; size = 4
$T4 = -12 ; size = 4
$T5 = -8 ; size = 4
$T6 = -4 ; size = 4
_main PROC
push ebp
mov ebp, esp
sub esp, 36 ; 00000024H
mov DWORD PTR _c$[ebp], 0 ; Myfirst
mov DWORD PTR _c$[ebp+4], 0 ; Mylast
mov DWORD PTR _c$[ebp+8], 0 ; Myend
lea eax, DWORD PTR _c$[ebp]
push eax
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
mov DWORD PTR $T6[ebp], 1
lea ecx, DWORD PTR $T6[ebp]
push ecx
lea ecx, DWORD PTR _c$[ebp]
call ?push_back@?$vector@HV?$allocator@H@std@@@std@@QAEX$$QAH@Z ; std::vector<int,std
::allocator<int> >::push_back
lea edx, DWORD PTR _c$[ebp]
push edx
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
mov DWORD PTR $T5[ebp], 2
lea eax, DWORD PTR $T5[ebp]
push eax
lea ecx, DWORD PTR _c$[ebp]
call ?push_back@?$vector@HV?$allocator@H@std@@@std@@QAEX$$QAH@Z ; std::vector<int,std
::allocator<int> >::push_back
lea ecx, DWORD PTR _c$[ebp]
push ecx
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
mov DWORD PTR $T4[ebp], 3
lea edx, DWORD PTR $T4[ebp]
push edx
lea ecx, DWORD PTR _c$[ebp]
call ?push_back@?$vector@HV?$allocator@H@std@@@std@@QAEX$$QAH@Z ; std::vector<int,std
::allocator<int> >::push_back
lea eax, DWORD PTR _c$[ebp]
push eax
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
mov DWORD PTR $T3[ebp], 4
lea ecx, DWORD PTR $T3[ebp]
push ecx
lea ecx, DWORD PTR _c$[ebp]
call ?push_back@?$vector@HV?$allocator@H@std@@@std@@QAEX$$QAH@Z ; std::vector<int,std
::allocator<int> >::push_back
lea edx, DWORD PTR _c$[ebp]
push edx
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
push 6
lea ecx, DWORD PTR _c$[ebp]
call ?reserve@?$vector@HV?$allocator@H@std@@@std@@QAEXI@Z ; std::vector<int,std::
allocator<int> >::reserve
lea eax, DWORD PTR _c$[ebp]
push eax
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
mov DWORD PTR $T2[ebp], 5
lea ecx, DWORD PTR $T2[ebp]
push ecx
lea ecx, DWORD PTR _c$[ebp]
call ?push_back@?$vector@HV?$allocator@H@std@@@std@@QAEX$$QAH@Z ; std::vector<int,std
::allocator<int> >::push_back
lea edx, DWORD PTR _c$[ebp]
push edx
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
mov DWORD PTR $T1[ebp], 6
lea eax, DWORD PTR $T1[ebp]
push eax
lea ecx, DWORD PTR _c$[ebp]
call ?push_back@?$vector@HV?$allocator@H@std@@@std@@QAEX$$QAH@Z ; std::vector<int,std
::allocator<int> >::push_back
lea ecx, DWORD PTR _c$[ebp]
push ecx
call ?dump@@YAXPAUvector_of_ints@@@Z ; dump
add esp, 4
push 5
lea ecx, DWORD PTR _c$[ebp]
call ?at@?$vector@HV?$allocator@H@std@@@std@@QAEAAHI@Z ; std::vector<int,std::
allocator<int> >::at
mov edx, DWORD PTR [eax]
push edx
push OFFSET $SG52650 ; ’%d’
call DWORD PTR __imp__printf
add esp, 8
mov eax, 8
shl eax, 2
mov ecx, DWORD PTR _c$[ebp]
mov edx, DWORD PTR [ecx+eax]
push edx
push OFFSET $SG52651 ; ’%d’
call DWORD PTR __imp__printf
add esp, 8
lea ecx, DWORD PTR _c$[ebp]
call ?_Tidy@?$vector@HV?$allocator@H@std@@@std@@IAEXXZ ; std::vector<int,std::
allocator<int> >::_Tidy
xor eax, eax
mov esp, ebp
pop ebp
ret 0
_main ENDP

我们看到.at() 方法如何进行辩解检查，当发生错误时将抛出异常。最后一个 printf() 调用只是从内存中读取数据，并没有做任何检查。 有人可能会问，为什么没有用类似 std::string 中 size 和 capacity 的变量，我猜想可能是为了让边界检查更快，但是我不确定。 GCC 生成的代码基本上相同，但是.at() 方法被内联了。

Listing34.12: GCC 4.8.1 -fno-inline-small-functions –O1

main proc near
push ebp
mov ebp, esp
push edi
push esi
push ebx
and esp, 0FFFFFFF0h
sub esp, 20h
mov dword ptr [esp+14h], 0
mov dword ptr [esp+18h], 0
mov dword ptr [esp+1Ch], 0
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov dword ptr [esp+10h], 1
lea eax, [esp+10h]
mov [esp+4], eax
lea eax, [esp+14h]
mov [esp], eax
call _ZNSt6vectorIiSaIiEE9push_backERKi ; std::vector<int,std::allocator<int
>>::push_back(int const&)
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov dword ptr [esp+10h], 2
lea eax, [esp+10h]
mov [esp+4], eax
lea eax, [esp+14h]
mov [esp], eax
call _ZNSt6vectorIiSaIiEE9push_backERKi ; std::vector<int,std::allocator<int
>>::push_back(int const&)
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov dword ptr [esp+10h], 3
lea eax, [esp+10h]
mov [esp+4], eax
lea eax, [esp+14h]
mov [esp], eax
call _ZNSt6vectorIiSaIiEE9push_backERKi ; std::vector<int,std::allocator<int
>>::push_back(int const&)
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov dword ptr [esp+10h], 4
lea eax, [esp+10h]
mov [esp+4], eax
lea eax, [esp+14h]
mov [esp], eax
call _ZNSt6vectorIiSaIiEE9push_backERKi ; std::vector<int,std::allocator<int
>>::push_back(int const&)
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov ebx, [esp+14h]
mov eax, [esp+1Ch]
sub eax, ebx
cmp eax, 17h
ja short loc_80001CF
mov edi, [esp+18h]
sub edi, ebx
sar edi, 2
mov dword ptr [esp], 18h
call _Znwj ; operator new(uint)
mov esi, eax
test edi, edi
jz short loc_80001AD
lea eax, ds:0[edi*4]
mov [esp+8], eax ; n
mov [esp+4], ebx ; src
mov [esp], esi ; dest
call memmove
loc_80001AD: ; CODE XREF: main+F8
mov eax, [esp+14h]
test eax, eax
jz short loc_80001BD
mov [esp], eax ; void *
call _ZdlPv ; operator delete(void *)
loc_80001BD: ; CODE XREF: main+117
mov [esp+14h], esi
lea eax, [esi+edi*4]
mov [esp+18h], eax
add esi, 18h
mov [esp+1Ch], esi
loc_80001CF: ; CODE XREF: main+DD
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov dword ptr [esp+10h], 5
lea eax, [esp+10h]
mov [esp+4], eax
lea eax, [esp+14h]
mov [esp], eax
call _ZNSt6vectorIiSaIiEE9push_backERKi ; std::vector<int,std::allocator<int
>>::push_back(int const&)
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov dword ptr [esp+10h], 6
lea eax, [esp+10h]
mov [esp+4], eax
lea eax, [esp+14h]
mov [esp], eax
call _ZNSt6vectorIiSaIiEE9push_backERKi ; std::vector<int,std::allocator<int
>>::push_back(int const&)
lea eax, [esp+14h]
mov [esp], eax
call _Z4dumpP14vector_of_ints ; dump(vector_of_ints *)
mov eax, [esp+14h]
mov edx, [esp+18h]
sub edx, eax
cmp edx, 17h
ja short loc_8000246
mov dword ptr [esp], offset aVector_m_range ; "vector::_M_range_check"
call _ZSt20__throw_out_of_rangePKc ; std::__throw_out_of_range(char const*)
loc_8000246: ; CODE XREF: main+19C
mov eax, [eax+14h]
mov [esp+8], eax
mov dword ptr [esp+4], offset aD ; "%d\n"
mov dword ptr [esp], 1
call __printf_chk
mov eax, [esp+14h]
mov eax, [eax+20h]
mov [esp+8], eax
mov dword ptr [esp+4], offset aD ; "%d\n"
mov dword ptr [esp], 1
call __printf_chk
mov eax, [esp+14h]
test eax, eax
jz short loc_80002AC
mov [esp], eax ; void *
call _ZdlPv ; operator delete(void *)
jmp short loc_80002AC
; ---------------------------------------------------------------------------
mov ebx, eax
mov edx, [esp+14h]
test edx, edx
jz short loc_80002A4
mov [esp], edx ; void *
call _ZdlPv ; operator delete(void *)
loc_80002A4: ; CODE XREF: main+1FE
mov [esp], ebx
call _Unwind_Resume
; ---------------------------------------------------------------------------
loc_80002AC: ; CODE XREF: main+1EA
; main+1F4
mov eax, 0
lea esp, [ebp-0Ch]
pop ebx
pop esi
pop edi
pop ebp
locret_80002B8: ; DATA XREF: .eh_frame:08000510
; .eh_frame:080005BC
retn
main endp

.reserve() 方法也被内联了。如果缓冲区大小小于新的 size，则调用 new() 申请新缓冲区，调用 memmove() 拷贝缓冲区内容，然后调用 delete() 释放旧的缓冲区。 让你给我们看看通过 GCC 编译后程序的输出

_Myfirst=0x(nil), _Mylast=0x(nil), _Myend=0x(nil)
size=0, capacity=0
_Myfirst=0x8257008, _Mylast=0x825700c, _Myend=0x825700c
size=1, capacity=1
element 0: 1
_Myfirst=0x8257018, _Mylast=0x8257020, _Myend=0x8257020
size=2, capacity=2
element 0: 1
element 1: 2
_Myfirst=0x8257028, _Mylast=0x8257034, _Myend=0x8257038
size=3, capacity=4
element 0: 1
element 1: 2
element 2: 3
_Myfirst=0x8257028, _Mylast=0x8257038, _Myend=0x8257038
size=4, capacity=4
element 0: 1
element 1: 2
element 2: 3
element 3: 4
_Myfirst=0x8257040, _Mylast=0x8257050, _Myend=0x8257058
size=4, capacity=6
element 0: 1
element 1: 2
element 2: 3
element 3: 4
_Myfirst=0x8257040, _Mylast=0x8257054, _Myend=0x8257058
size=5, capacity=6
element 0: 1
element 1: 2
element 2: 3
element 3: 4
element 4: 5
_Myfirst=0x8257040, _Mylast=0x8257058, _Myend=0x8257058
size=6, capacity=6
element 0: 1
element 1: 2
element 2: 3
element 3: 4
element 4: 5
element 5: 6
6
0

我们可以看到缓冲区大小的增长不同于 MSVC 中。 简单的实验说明 MSVC 中缓冲区每次扩大当前大小的 50%，而 GCC 中则每次扩大 100%。