问题描述
为什么这两种情况（A和C的初始化）在C ++ 14中产生不同的默认初始化结果？
基于cppreference.com中的默认初始化规则，我无法理解结果
Why those two scenarios (the initialization of A and of C ) produce different default initialization results in C++ 14?I can't understand the result based on the default initialization rules in cppreference.com
struct A { int m; };
struct C { C() : m(){}; int m; };
int main() {
A *a, *d;
A b;
A c{};
a=new A();
d=new A;
cout<<a->m<<endl;
cout<<d->m<<endl;
cout<<b.m<<endl;
cout<<c.m<<endl;
cout<<"--------------------"<<endl;
C *a1, *d1;
C b1;
C c1{};
a1=new C();
d1=new C;
cout<<a1->m<<endl;
cout<<d1->m<<endl;
cout<<b1.m<<endl;
cout<<c1.m<<endl;
}
输出：
(Scenario 1)
0
-1771317376
-1771317376
0
--------------------
(Scenario 2)
0
0
0
0
试图解释这一点的帖子（但是我还不清楚为什么结果的差异以及在每种情况下导致m初始化的原因）：
The post that tries to explain this (but it's not clear to me yet why the difference in the results and what causes m to be initialized in each scenario): Default, value and zero initialization mess
推荐答案
A 没有用户定义的构造函数，因此生成了默认构造函数。 C 具有用户定义的构造函数，因此不能保证会生成默认的构造函数，尤其是由于用户定义的构造函数会重载默认的构造函数。几乎可以肯定，C的每个构造都使用用户定义的构造函数。
A has no user defined constructors, so a default constructor was generated. C has a user defined constructor, so there is no guarantee that a default constructor was generated, especially since the user defined constructor overloads the default constructor. It is almost certain that every construction of C used the user defined constructor.
在用户定义的 C 构造函数中使用初始化列表对 C :: m 进行值初始化。初始化 C :: m 时，将对其进行值初始化，其中包括零初始化。
In the user defined constructor for C uses an initialization list to value initialize C::m. When C::m is initialized, it is value initialized, which includes a zero initialization.
new A; 和 A b; 是默认初始化。这不会设置为其成员分配任何值。 A :: m 中存储的值是未定义的行为。
new A; and A b; are default initializations. This does not set assign any values to their members. What value is stored in A::m is undefined behavior.
new A（ ）； 和 A c {}; 是值初始化。作为值初始化的一部分，它将执行零初始化。
new A(); and A c{}; are value initializations. As part of the value initialization, it performs a zero initialization.
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