#include <iostream> using namespace std; class Base1 { protected: int data1; public: Base1(int a=8) { data1 = a; cout<<data1<<", Base1 Constructor\n"; } ~Base1( ) { cout<<data1<<", Base1 Destructor\n"; } }; class Base2 { protected: int data2; public: Base2(int a=9) { data2 = a; cout<<data2<<", Base2 Constructor\n"; } ~Base2( ) { cout<<data2<<", Base2 Destructor\n"; } }; class Derived:public Base1, public Base2 { int d; public: Derived(int x,int y,int z) //A { d=z; cout<<"Derived Constructor\n"; } ~Derived( ) { cout<<"Derived Destructor\n"; } void Show( ) { cout<<data1<<','<<data2<<','<<d<<endl; } }; int main( ) { Derived c(1, 2, 3); c.Show( ); return 0; }分析对象的构造函数和析构函数调用时机

在这个程序中，我们定义了三个类：Base1、Base2和Derived。Derived类继承自Base1和Base2类。

在创建一个Derived对象时，会调用Derived类的构造函数。在构造函数中，我们可以看到，我们定义了一个整数d并将其赋值为z，然后我们输出了“Derived Constructor”。

在Derived类的构造函数中，我们还没有初始化Base1和Base2类。因此，在构造Derived对象时，程序会自动调用Base1和Base2类的默认构造函数。Base1类的构造函数将data1初始化为8，并输出“8, Base1 Constructor”。Base2类的构造函数将data2初始化为9，并输出“9, Base2 Constructor”。

在调用Derived类的析构函数时，程序会先调用Derived类的析构函数，然后分别调用Base2和Base1类的析构函数。因此，我们会看到输出顺序为“Derived Destructor”、“9, Base2 Destructor”和“8, Base1 Destructor”。

在main函数中，我们创建了一个Derived对象c并调用了它的Show函数。在Show函数中，我们输出了Base1和Base2类的成员变量data1和data2的值，以及Derived类的成员变量d的值。因此，程序会输出“8,9,3”。

相关问题

#include <iostream> using namespace std; class Base1 { protected: int data1; public: Base1(int a = 8) { data1 = a; cout << data1 << ", Base1 Constructor\n"; } ~Base1() { cout << data1 << ", Base1 Destructor\n"; } }; class Base2 { protected: int data2; public: Base2(int a = 9) { data2 = a; cout << data2 << ", Base2 Constructor\n"; } ~Base2() { cout << data2 << ", Base2 Destructor\n"; } }; class Derived :public Base1, public Base2 { int d; public: Derived(int x, int y, int z) : Base1(x), Base2(y) //A { d = z; cout << "Derived Constructor\n"; } ~Derived() { cout << "Derived Destructor\n"; } void Show() { cout << data1 << ',' << data2 << ',' << d << endl; } }; int main() { Derived c(1, 2, 3); c.Show(); return 0; }以上C++代码调试时显示有一个错误，怎么解决？

在代码的构造函数`Derived(int x, int y, int z) : Base1(x), Base2(y)`后面没有加上大括号`{}`，导致编译器无法正确识别构造函数的定义，因此报错。

正确的代码应该是：

#include <iostream>
using namespace std;

class Base1 {
protected:
    int data1;
public:
    Base1(int a = 8) {
        data1 = a;
        cout << data1 << ", Base1 Constructor\n";
    }
    ~Base1() {
        cout << data1 << ", Base1 Destructor\n";
    }
};

class Base2 {
protected:
    int data2;
public:
    Base2(int a = 9) {
        data2 = a;
        cout << data2 << ", Base2 Constructor\n";
    }
    ~Base2() {
        cout << data2 << ", Base2 Destructor\n";
    }
};

class Derived :public Base1, public Base2 {
    int d;
public:
    Derived(int x, int y, int z) : Base1(x), Base2(y) {
        d = z;
        cout << "Derived Constructor\n";
    }
    ~Derived() {
        cout << "Derived Destructor\n";
    }
    void Show() {
        cout << data1 << ',' << data2 << ',' << d << endl;
    }
};

int main() {
    Derived c(1, 2, 3);
    c.Show();
    return 0;
}

以上代码中，在`Derived`类定义后面加上了大括号`{}`，用于定义构造函数和析构函数的内容。这样就可以正确识别构造函数的定义，消除了编译器报错。

运行代码，进入调试模式，分析对象的构造函数和析构函数调用时机，代码如下： #include <iostream> using namespace std; class Base1 { protected: int data1; public: Base1(int a=8) { data1 = a; cout<<data1<<", Base1 Constructor\n"; } ~Base1( ) { cout<<data1<<", Base1 Destructor\n"; } }; class Base2 { protected: int data2; public: Base2(int a=9) { data2 = a; cout<<data2<<", Base2 Constructor\n"; } ~Base2( ) { cout<<data2<<", Base2 Destructor\n"; } }; class Derived:public Base1, public Base2 { int d; public: Derived(int x, int y, int z) : Base1(x), Base2(y) //A { d=z; cout<<"Derived Constructor\n"; } ~Derived( ) { cout<<"Derived Destructor\n"; } void Show( ) { cout<<data1<<','<<data2<<','<<d<<endl; } }; int main( ) { Derived c(1, 2, 3); c.Show( ); return 0; }先分析结果然后如果将A行代码替换为 Derived(int x,int y,int z)，分析程序结果

程序的输出结果是：

1, Base1 Constructor
2, Base2 Constructor
Derived Constructor
1,2,3
Derived Destructor
2, Base2 Destructor
1, Base1 Destructor

如果将A行代码替换为`Derived(int x,int y,int z)`，输出结果不会有变化，因为这是派生类的构造函数，会自动调用基类的构造函数。