C++中, 构造函数和析构函数能不能被显示调用?

代码:

#include <iostream>using namespace std;class A{public:    A()    {        cout << "Default constructor is called./r/n";    }    A(int ix)    {        cout << "Another constructor is called./r/n";    }    ~A()    {        cout << "Destructor is called./r/n";    }};int main(){    A a1;          // <1>    a1.A::A();     // <2> 显示调用默认构造函数(写成a1.A()会报错)    a1.A::A(7);    // <3> 显示调用非默认构造函数(写成a1.A(7)会报错)    a1.A::~A();    // <4> 显示调用析构函数, 但是此时对象a1并没有销毁(写成a1.~A()不会报错)    // A a2();     // 这样写没报错, 但也没调用任何构造函数和析构函数.    A a3 = A();    // <5> 完整写法: A a3 = A::A();    A a4(77);      // <6>    A a5 = A(777); // <7> 完整写法: A a5 = A::A(777);    return 0;    // <8>, <9>, <10>, <11> return语句之后, 右括号之前析构函数被隐式调用. a1, a3, a4, a5对象在这里被销毁.}#include <iostream>using namespace std;class </span></span><br /><hr><link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/bootstrap/5.2.1/css/bootstrap.min.css" integrity="sha512-siwe/oXMhSjGCwLn+scraPOWrJxHlUgMBMZXdPe2Tnk3I0x3ESCoLz7WZ5NTH6SZrywMY+PB1cjyqJ5jAluCOg==" crossorigin="anonymous" referrerpolicy="no-referrer" />
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