C++ 构造函数答疑

#include <string>
#include <iostream>

using namespace std;

class Base
{
    public:
        Base(){
            cout<<"Base Construct.\n";
        }
        Base(const Base& b){
            cout<<"Copy Construct of Base.\n";
        }
        virtual void Print(int i=0){
            cout<<"Base\t"<<i<<endl;
        }
};

class Derive:public Base
{
    public:
        Derive(){
            cout<<"Derive Construct.\n";
        }
        void Print(int i=1){
            cout<<"Derive\t"<<i<<endl;
        }
};

void B1(Base b)
{
    b.Print();
}

void B2(Base &b)
{
    b.Print();
}
void B3(Base *b)
{
    b->Print();
}
void D1(Derive d)
{
    d.Print();
}
void D2(Derive &d)
{
    d.Print();
}
void D3(Derive *d)
{
    d->Print();
}
void drawLine()
{
    cout<<"---------------\n";
}

int main()
{
    Base b;
    drawLine();
    Derive d;
    drawLine();
    Base* pb = new Derive{};
    drawLine();
    B1(d);
    drawLine();
    B2(d);
    drawLine();
    B3(&d);
    drawLine();
    D1(d);
    drawLine();
    D2(d);
    drawLine();
    D3(&d);
}

上述代码的输出为

Base Construct.
---------------
Base Construct.
Derive Construct.
---------------
Base Construct.
Derive Construct.
---------------
Copy Construct of Base.
Base    0
---------------
Derive  0
---------------
Derive  0
---------------
Copy Construct of Base.
Derive  1
---------------
Derive  1
---------------
Derive  1

得出的结论如下：

• 子类的构造函数一定会先构造父类
• When an object as a parameter passed-by-value, it will call Copy Constructor
• When an object as a parameter passed-by-reference, it has none business with constructor. 按引用传参和其指针效果一样。
• 子类可以传给参数是父类的函数，会调用父类的复制构造函数。该参数实则已经是父类的对象，调用的虚函数也是父类虚函数表里的，参数的默认值也是父类的。
• 虚函数的默认值跟随虚函数表走的。也就是上述Print(int i)中i的默认值取决于对象的虚函数表(是父类的还是子类的)。