以下是多重继承中遇到的钻石问题,
class Base {
public:
Base() {
cout << "Empty Base constructor " << endl;
}
Base(const string & strVar) {
m_strVar = strVar;
cout << m_strVar << endl;
}
virtual ~Base() {
cout << "Empty Base destructor " << endl;
}
virtual const string & strVar() const {
return m_strVar;
}
string m_strVar;
};
class Derived1: public virtual Base {
public:
Derived1() {
cout << "Empty Derived1 constructor " << endl;
}
Derived1(const string & strVar) : Base(strVar) {
cout << " Derived1 one arg constructor" << endl;
}
~Derived1() {
cout << "Empty Derived1 destructor " << endl;
}
};
class Derived2: public virtual Base {
public:
Derived2() {
cout << "Empty Derived2 constructor " << endl;
}
Derived2(const string & strVar) : Base(strVar) {
cout << "Derived2 one arg constructor" << endl;
}
~Derived2() {
cout << "Empty Derived2 destructor " << endl;
}
};
class Derived: public Derived1, public Derived2 {
public:
Derived(const string & strVar) : Derived1(strVar), Derived2(strVar) {
cout << "Derived Constructor " << endl;
}
~Derived() {
cout << "Empty Derived destructor " << endl;
}
};
int main() {
Derived derObj ("Print this if you can ! ");
}
我得到的输出是
我想知道为什么我的derObj参数,即“如果可以的话打印”不打印,输出不像
答案 0 :(得分:6)
这与虚拟继承有关。
当一个类被虚拟继承时,层次结构中大多数派生类的责任就是调用它的构造函数:here Derived
。
由于Base
是默认可构造的,并且您没有精确的任何内容,Derived
会调用Base
的默认构造函数。
如果要打印字符串,请使用:
Derived(const string & strVar) : Base(strVar), Derived1(strVar), Derived2(strVar)
{
std::cout << "Derived Constructor\n";
}
您可以删除默认构造函数以让编译器诊断问题,但并非所有编译器都提供非常有用的消息。