new和malloc关键字
数据结构中有一个问题，链表后插。
这个操作需要新建节点，自然要用到new或者malloc关键字。c中较为严谨，在malloc后需要判断是否能新建节点成功（LNode2 *s2=(LNode2*) malloc(sizeof(LNode2));），当然99.99%是能成功的，除非计算机内存实在不够了...
于是临下班前试着用c++实现一遍，换成类来描述，这个判断自然就变成了
LNode *S = new LNode();
if (S == nullptr)
return false;//内存不足分配失败问题就出在这里了。IDE(Clion)提示我"Condition is always false"，并且VS好像还不会报出来这个。
因此下来研究、学习了一下。C++中operator new的行为及其与malloc的区别。new在实现中会调用malloc并且由编译器安插调用构造函数的代码，malloc失败返回0，而new直接抛异常。但不清楚有没有不抛异常的new和不调用构造函数的new这一点。下面结合代码分析一下。
19: CTest *pTest = new CTest(3, 2);
0027652D 6A 08 push 8
0027652F E8 8D AD FF FF call operator new (02712C1h) ; 默认的operator new
00276534 83 C4 04 add esp,4
00276537 89 85 08 FF FF FF mov dword ptr [ebp-0F8h],eax
0027653D C7 45 FC 00 00 00 00 mov dword ptr [ebp-4],0
00276544 83 BD 08 FF FF FF 00 cmp dword ptr [ebp-0F8h],0
0027654B 74 17 je main+74h (0276564h)
0027654D 6A 02 push 2
0027654F 6A 03 push 3
00276551 8B 8D 08 FF FF FF mov ecx,dword ptr [ebp-0F8h]
00276557 E8 B4 AE FF FF call CTest::CTest (0271410h) ; 调用构造函数
0027655C 89 85 E8 FE FF FF mov dword ptr [ebp-118h],eax
00276562 EB 0A jmp main+7Eh (027656Eh)
00276564 C7 85 E8 FE FF FF 00 00 00 00 mov dword ptr [ebp-118h],0
0027656E 8B 85 E8 FE FF FF mov eax,dword ptr [ebp-118h]
00276574 89 85 14 FF FF FF mov dword ptr [ebp-0ECh],eax
0027657A C7 45 FC FF FF FF FF mov dword ptr [ebp-4],0FFFFFFFFh
00276581 8B 8D 14 FF FF FF mov ecx,dword ptr [ebp-0ECh]
00276587 89 4D EC mov dword ptr [pTest],ecx
20:
21: CTest *pTest2 = new (std::nothrow) CTest(4, 6);
0027658A 68 D0 C2 27 00 push offset std::nothrow (027C2D0h)
0027658F 6A 08 push 8
00276591 E8 63 AC FF FF call operator new (02711F9h) ;多个参数，不抛异常的operator new
00276596 83 C4 08 add esp,8
00276599 89 85 F0 FE FF FF mov dword ptr [ebp-110h],eax
0027659F C7 45 FC 01 00 00 00 mov dword ptr [ebp-4],1
002765A6 83 BD F0 FE FF FF 00 cmp dword ptr [ebp-110h],0
002765AD 74 17 je main+0D6h (02765C6h)
002765AF 6A 06 push 6
002765B1 6A 04 push 4
002765B3 8B 8D F0 FE FF FF mov ecx,dword ptr [ebp-110h]
002765B9 E8 52 AE FF FF call CTest::CTest (0271410h) ; 调用构造函数
002765BE 89 85 E8 FE FF FF mov dword ptr [ebp-118h],eax
20:
21: CTest *pTest2 = new (std::nothrow) CTest(4, 6);
002765C4 EB 0A jmp main+0E0h (02765D0h)
002765C6 C7 85 E8 FE FF FF 00 00 00 00 mov dword ptr [ebp-118h],0
002765D0 8B 85 E8 FE FF FF mov eax,dword ptr [ebp-118h]
002765D6 89 85 FC FE FF FF mov dword ptr [ebp-104h],eax
002765DC C7 45 FC FF FF FF FF mov dword ptr [ebp-4],0FFFFFFFFh
002765E3 8B 8D FC FE FF FF mov ecx,dword ptr [ebp-104h]
002765E9 89 4D E0 mov dword ptr [pTest2],ecx
22:
23: pTest->m_dw1 = 3;
002765EC 8B 45 EC mov eax,dword ptr [pTest]
002765EF C7 00 03 00 00 00 mov dword ptr [eax],3
24: pTest2->m_dw2 = 5;
002765F5 8B 45 E0 mov eax,dword ptr [pTest2]
002765F8 C7 40 04 05 00 00 00 mov dword ptr [eax+4],5
25:
26:
27: return 0;
002765FF 33 C0 xor eax,eax
28: }operator new的源码：
void* __CRTDECL operator new(size_t const size)
{
for (;;)
{
if (void* const block = malloc(size))
{
return block;
}
if (_callnewh(size) == 0)
{
if (size == SIZE_MAX)
{
__scrt_throw_std_bad_array_new_length();
}
else
{
__scrt_throw_std_bad_alloc();
}
}
// The new handler was successful; try to allocate again...
}
}operator new的源码，嗯嗯，果然不可能抛异常，因为异常都被接住了：
void* __CRTDECL operator new(size_t const size, std::nothrow_t const&) noexcept
{
try
{
return operator new(size);
}
catch (...)
{
return nullptr;
}
}delete操作都是一样的，调用了带两个参数的operator delete，这个应该也是后来搞出来的一个函数，我印象还比较深。之前在尝试将VS2022编译的obj与WDK中lib版CRT库匹配时曾遇到过这个带了两个参数的delete找不到匹配的符号，结果用了/Zc:sizedDealloc-开关才避免其生成。
26: delete(pTest);
013865FF 8B 45 EC mov eax,dword ptr [pTest]
01386602 89 85 E4 FE FF FF mov dword ptr [ebp-11Ch],eax
01386608 6A 08 push 8
0138660A 8B 8D E4 FE FF FF mov ecx,dword ptr [ebp-11Ch]
01386610 51 push ecx
01386611 E8 3F AA FF FF call operator delete (01381055h)
01386616 83 C4 08 add esp,8
27: delete(pTest2);
01386619 8B 45 E0 mov eax,dword ptr [pTest2]
0138661C 89 85 D8 FE FF FF mov dword ptr [ebp-128h],eax
01386622 6A 08 push 8
01386624 8B 8D D8 FE FF FF mov ecx,dword ptr [ebp-128h]
0138662A 51 push ecx
0138662B E8 25 AA FF FF call operator delete (01381055h)
01386630 83 C4 08 add esp,8 在其中调用了带一个参数的operator delete：
28: void __CRTDECL operator delete(void* const block, size_t const) noexcept
29: {
01384FF0 55 push ebp
01384FF1 8B EC mov ebp,esp
30: operator delete(block);
01384FF3 8B 45 08 mov eax,dword ptr [block]
30: operator delete(block);
01384FF6 50 push eax
01384FF7 E8 36 C0 FF FF call operator delete (01381032h)
01384FFC 83 C4 04 add esp,4
31: }
01384FFF 5D pop ebp
01385000 C3 ret 结论：可以通过std::nothrow让new不抛异常，VS22下的做法是在该new函数中用try捕获了异常，并不再抛出，同时让其返回0；对于有构造函数的类对象，new时都会调用其构造函数，是由编译器自动插入代码的。
最最后，JetBrains真香。