Question

MSVC 实现形式基本相同，但是它存储了当前 list 的大小。这意味着.size() 方法非常迅速，只需要从内存中读取一个值即可。从另一个方面来说，必须在插入和删除操作后更新 size 变量的值。 MSVC 组织节点的方式也稍有不同。

GCC 将垃圾元素放在链表的尾部，而 MSVC 放在链表起始位置。

Listing 34.9: MSVC 2012 /Fa2.asm /Ox /GS- /Ob1

_l$ = -16 ; size = 8
_t1$ = -8 ; size = 8
_main PROC
sub esp, 16 ; 00000010H
push ebx
push esi
push edi
push 0
push 0
lea ecx, DWORD PTR _l$[esp+36]
mov DWORD PTR _l$[esp+40], 0
; allocate first "garbage" element
call ?_Buynode0@?$_List_alloc@$0A@U?$_List_base_types@Ua@@V?
$allocator@Ua@@@std@@@std@@@std@@QAEPAU?$_List_node@Ua@@PAX@2@PAU32@0@Z ; std::_List_alloc<0,
std::_List_base_types<a,std::allocator<a> > >::_Buynode0
mov edi, DWORD PTR __imp__printf
mov ebx, eax
push OFFSET $SG40685 ; ’* empty list:’
mov DWORD PTR _l$[esp+32], ebx
call edi ; printf
lea eax, DWORD PTR _l$[esp+32]
push eax
call ?dump_List_val@@YAXPAI@Z ; dump_List_val
mov esi, DWORD PTR [ebx]
add esp, 8
lea eax, DWORD PTR _t1$[esp+28]
push eax
push DWORD PTR [esi+4]
lea ecx, DWORD PTR _l$[esp+36]
push esi
mov DWORD PTR _t1$[esp+40], 1 ; data for a new node
mov DWORD PTR _t1$[esp+44], 2 ; data for a new node
; allocate new node
call ??$_Buynode@ABUa@@@?$_List_buy@Ua@@V?$allocator@Ua@@@std@@@std@@QAEPAU?
$_List_node@Ua@@PAX@1@PAU21@0ABUa@@@Z ; std::_List_buy<a,std::allocator<a> >::_Buynode<a
const &>
mov DWORD PTR [esi+4], eax
mov ecx, DWORD PTR [eax+4]
mov DWORD PTR _t1$[esp+28], 3 ; data for a new node
mov DWORD PTR [ecx], eax
mov esi, DWORD PTR [ebx]
lea eax, DWORD PTR _t1$[esp+28]
push eax
push DWORD PTR [esi+4]
lea ecx, DWORD PTR _l$[esp+36]
push esi
mov DWORD PTR _t1$[esp+44], 4 ; data for a new node
; allocate new node
call ??$_Buynode@ABUa@@@?$_List_buy@Ua@@V?$allocator@Ua@@@std@@@std@@QAEPAU?
$_List_node@Ua@@PAX@1@PAU21@0ABUa@@@Z ; std::_List_buy<a,std::allocator<a> >::_Buynode<a
const &>
mov DWORD PTR [esi+4], eax
mov ecx, DWORD PTR [eax+4]
mov DWORD PTR _t1$[esp+28], 5 ; data for a new node
mov DWORD PTR [ecx], eax
lea eax, DWORD PTR _t1$[esp+28]
push eax
push DWORD PTR [ebx+4]
lea ecx, DWORD PTR _l$[esp+36]
push ebx
mov DWORD PTR _t1$[esp+44], 6 ; data for a new node
; allocate new node
call ??$_Buynode@ABUa@@@?$_List_buy@Ua@@V?$allocator@Ua@@@std@@@std@@QAEPAU?
$_List_node@Ua@@PAX@1@PAU21@0ABUa@@@Z ; std::_List_buy<a,std::allocator<a> >::_Buynode<a
const &>
mov DWORD PTR [ebx+4], eax
mov ecx, DWORD PTR [eax+4]
push OFFSET $SG40689 ; ’* 3-elements list:’
mov DWORD PTR _l$[esp+36], 3
mov DWORD PTR [ecx], eax
call edi ; printf
lea eax, DWORD PTR _l$[esp+32]
push eax
call ?dump_List_val@@YAXPAI@Z ; dump_List_val
push OFFSET $SG40831 ; ’node at .begin:’
call edi ; printf
push DWORD PTR [ebx] ; get next field of node $l$ variable points to
call ?dump_List_node@@YAXPAUList_node@@@Z ; dump_List_node
push OFFSET $SG40835 ; ’node at .end:’
call edi ; printf
push ebx ; pointer to the node $l$ variable points to!
call ?dump_List_node@@YAXPAUList_node@@@Z ; dump_List_node
push OFFSET $SG40839 ; ’* let’’s count from the begin:’
call edi ; printf
mov esi, DWORD PTR [ebx] ; operator++: get ->next pointer
push DWORD PTR [esi+12]
push DWORD PTR [esi+8]
push OFFSET $SG40846 ; ’1st element: %d %d’
call edi ; printf
mov esi, DWORD PTR [esi] ; operator++: get ->next pointer
push DWORD PTR [esi+12]
push DWORD PTR [esi+8]
push OFFSET $SG40848 ; ’2nd element: %d %d’
call edi ; printf
mov esi, DWORD PTR [esi] ; operator++: get ->next pointer
push DWORD PTR [esi+12]
push DWORD PTR [esi+8]
push OFFSET $SG40850 ; ’3rd element: %d %d’
call edi ; printf
mov eax, DWORD PTR [esi] ; operator++: get ->next pointer
add esp, 64 ; 00000040H
push DWORD PTR [eax+12]
push DWORD PTR [eax+8]
push OFFSET $SG40852 ; ’element at .end(): %d %d’
call edi ; printf
push OFFSET $SG40853 ; ’* let’’s count from the end:’
call edi ; printf
push DWORD PTR [ebx+12] ; use x and y fields from the node $l$ variable points to
push DWORD PTR [ebx+8]
push OFFSET $SG40860 ; ’element at .end(): %d %d’
call edi ; printf
mov esi, DWORD PTR [ebx+4] ; operator--: get ->prev pointer
push DWORD PTR [esi+12]
push DWORD PTR [esi+8]
push OFFSET $SG40862 ; ’3rd element: %d %d’
call edi ; printf
mov esi, DWORD PTR [esi+4] ; operator--: get ->prev pointer
push DWORD PTR [esi+12]
push DWORD PTR [esi+8]
push OFFSET $SG40864 ; ’2nd element: %d %d’
call edi ; printf
mov eax, DWORD PTR [esi+4] ; operator--: get ->prev pointer
push DWORD PTR [eax+12]
push DWORD PTR [eax+8]
push OFFSET $SG40866 ; ’1st element: %d %d’
call edi ; printf
add esp, 64 ; 00000040H
push OFFSET $SG40867 ; ’removing last element...’
call edi ; printf
mov edx, DWORD PTR [ebx+4]
add esp, 4
; prev=next?
; it is the only element, "garbage one"?
; if yes, do not delete it!
cmp edx, ebx
je SHORT $LN349@main
mov ecx, DWORD PTR [edx+4]
mov eax, DWORD PTR [edx]
mov DWORD PTR [ecx], eax
mov ecx, DWORD PTR [edx]
mov eax, DWORD PTR [edx+4]
push edx
mov DWORD PTR [ecx+4], eax
call ??3@YAXPAX@Z ; operator delete
add esp, 4
mov DWORD PTR _l$[esp+32], 2
$LN349@main:
lea eax, DWORD PTR _l$[esp+28]
push eax
call ?dump_List_val@@YAXPAI@Z ; dump_List_val
mov eax, DWORD PTR [ebx]
add esp, 4
mov DWORD PTR [ebx], ebx
mov DWORD PTR [ebx+4], ebx
cmp eax, ebx
je SHORT $LN412@main
$LL414@main:
mov esi, DWORD PTR [eax]
push eax
call ??3@YAXPAX@Z ; operator delete
add esp, 4
mov eax, esi
cmp esi, ebx
jne SHORT $LL414@main
$LN412@main:
push ebx
call ??3@YAXPAX@Z ; operator delete
add esp, 4
xor eax, eax
pop edi
pop esi
pop ebx
add esp, 16 ; 00000010H
ret 0
_main ENDP

与 GCC 不同，MSVC 代码在函数开始，通过 Buynode 函数分配垃圾元素，这个方法也用于后续节点的申请（GCC 将最早的节点分配在栈上）。

Listing 34.10 完整输出

* empty list:
_Myhead=0x003CC258, _Mysize=0
ptr=0x003CC258 _Next=0x003CC258 _Prev=0x003CC258 x=6226002 y=4522072
* 3-elements list:
_Myhead=0x003CC258, _Mysize=3
ptr=0x003CC258 _Next=0x003CC288 _Prev=0x003CC2A0 x=6226002 y=4522072
ptr=0x003CC288 _Next=0x003CC270 _Prev=0x003CC258 x=3 y=4
ptr=0x003CC270 _Next=0x003CC2A0 _Prev=0x003CC288 x=1 y=2
ptr=0x003CC2A0 _Next=0x003CC258 _Prev=0x003CC270 x=5 y=6
node at .begin:
ptr=0x003CC288 _Next=0x003CC270 _Prev=0x003CC258 x=3 y=4
node at .end:
ptr=0x003CC258 _Next=0x003CC288 _Prev=0x003CC2A0 x=6226002 y=4522072
* let’s count from the begin:
1st element: 3 4
2nd element: 1 2
3rd element: 5 6
element at .end(): 6226002 4522072
* let’s count from the end:
element at .end(): 6226002 4522072
3rd element: 5 6
2nd element: 1 2
1st element: 3 4
removing last element...
_Myhead=0x003CC258, _Mysize=2
ptr=0x003CC258 _Next=0x003CC288 _Prev=0x003CC270 x=6226002 y=4522072
ptr=0x003CC288 _Next=0x003CC270 _Prev=0x003CC258 x=3 y=4
ptr=0x003CC270 _Next=0x003CC258 _Prev=0x003CC288 x=1 y=2