問(wèn)題：
A是1個(gè)類(lèi)，我們常常new1個(gè)A的對(duì)象，你常常會(huì)這么寫(xiě)：
A *pa = new A();但是如果我這么寫(xiě)呢：
A *pa = new A;二者有甚么區(qū)分嗎？ 這就是我們今天要討論的問(wèn)題。
先給出答案，再漸漸解釋
If the class has a default constructor defined, then both are equivalent; the object will be created by calling that constructor.
If the class only has an implicit default constructor, then there is a difference. The first will leave any members of POD type uninitialised; the second will value-initialise them .
甚么是POD
POD的意思是 Plain Old Data，即1個(gè)class或struct沒(méi)有構(gòu)造函數(shù)、析構(gòu)函數(shù)、虛函數(shù)。維基百科上這么描寫(xiě)：
A Plain Old Data Structure in C++ is an aggregate class that contains only PODS as members, has no user-defined destructor, no user-defined copy assignment operator, and no nonstatic members of pointer-to-member type.int, char, wchar_t, bool, float, double are PODs, as are long/short and signed/unsigned versions of them.
pointers (including pointer-to-function and pointer-to-member) are PODs,
enums are PODs
a const or volatile POD is a POD.
a class, struct or union of PODs is a POD provided that all non-static data members are public, and it has no base class and no constructors, destructors, or virtual methods. Static members don’t stop something being a POD under this rule.
我們就定義1個(gè)POD的類(lèi)：
class A
{
int m;
};再定義兩個(gè)非POD類(lèi)：
class B
{
~B();
};
class C
{
C() : m() {};
int m;
}; 接下來(lái)就是使用了：
#include<iostream>
using namespace std;
class A
{
public:
int m;
};
class B
{
public:
~B() {};
public:
int m;
};
class C
{
public:
C() : m() {};
public:
int m;
};
int main()
{
A *aObj1 = new A;
A *aObj2 = new A();
cout << aObj1->m << endl;
cout << aObj2->m << endl;
B *bObj1 = new B;
B *bObj2 = new B();
cout << bObj1->m << endl;
cout << bObj2->m << endl;
C *cObj1 = new C;
C *cObj2 = new C();
cout << cObj1->m << endl;
cout << cObj2->m << endl;
delete aObj1;
delete aObj2;
delete bObj1;
delete bObj2;
delete cObj1;
delete cObj2;
return 0;
}輸出：
18780112
0
4522067
0
0
0
在所有C++版本中，只有當(dāng)A是POD類(lèi)型的時(shí)候，new A和new A()才會(huì)有區(qū)分
就是再看看如何判斷是否是POD，不用認(rèn)為判斷：
#include <iostream>
#include <type_traits>
struct A {
int m;
};
struct B {
int m1;
private:
int m2;
};
struct C {
virtual void foo();
};
int main()
{
std::cout << std::boolalpha;
std::cout << std::is_pod<A>::value << '\n';
std::cout << std::is_pod<B>::value << '\n';
std::cout << std::is_pod<C>::value << '\n';
}
輸出：
true
false
false
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