问题描述
我的问题与有些重叠，几个其他类似的。这些有一些很好的答案，但我已经阅读他们，我仍然困惑，所以请不要考虑这个问题一式两份。
My question somewhat overlaps with this and several other similar ones. Those have some great answers, but I've read them and I'm still confused, so please don't consider this question a duplicate.
所以，我有以下代码：
class A {
public: int _a;
}
void main()
{
A inst1;
A* inst2 = new A;
A* inst3 = new A();
}
未初始化在 inst1 和 inst2 中，并初始化为 0 inst3 。
调用哪个初始化，为什么代码工作？请考虑我的帐户我手头没有C ++ 03标准，但我有最后的C ++ 11草稿（我编程的'03标准，虽然），所以引用'03标准或引用'11是最受欢迎的。
_a is left uninitialized in inst1 and inst2 and is initialized to 0 in inst3.Which initialization is called which, and why does the code work as it does? Please take into I account I don't have a C++ 03 standard at hand, but I have the last C++ 11 draft (I'm programming by '03 standard, though), so quotations of the '03 standard or references to '11 are most welcome.
P。这个研究的原始任务是正确地初始化任意模板类型 T 的成员。
P. S. The original task behind this research was to correctly zeto-initialize a member of arbitrary template type T.
推荐答案
不难：
A x;
A * p = new A;
这两个是默认初始化。因为你没有用户定义的构造函数，这只是意味着所有成员都是默认初始化的。默认初始化基本类型如 int 表示无初始化。
These two are default initialization. Since you don't have a user-defined constructor, this just means that all members are default-initialized. Default-initializing a fundamental type like int means "no initialization".
下一页：
A * p = new A();
这是值初始化。 （我不认为在C ++ 98/03有一个自动版本，虽然在C ++ 11你可以说 A x {}; 此外， A x = A（）; 实际上足够接近，尽管是复制初始化或 A x（（A（）））。
This is value initialization. (I don't think there exists an automatic version of this in C++98/03, though in C++11 you can say A x{};, and this brace-initialization becomes value-initialization. Moreover, A x = A(); is close enough practically despite being copy-initialization, or A x((A())) despite being direct-initialization.)
同样，在你的情况下，这意味着所有成员都是值初始化的。基本类型的值初始化意味着零初始化，这反过来意味着变量被初始化为零（所有基本类型都具有）。
Again, in your case this just means that all members are value-initialized. Value initialization for fundamental types means zero-initialization, which in turn means that the variables are initialized to zero (which all fundamental types have).
对于类类型的对象，default和initialize都调用默认构造函数。然后发生什么取决于构造函数的初始化器列表，并且游戏继续递归的成员变量。
For objects of class type, both default- and value-initialization invoke the default constructor. What happens then depends on the constructor's initializer list, and the game continues recursively for member variables.
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