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标题: 给基类构造函数传递参数 [打印本页]

作者: beyes 时间: 2011-8-28 00:23
标题: 给基类构造函数传递参数
给基类中的构造函数传递参数的通用形式如下：

derived-constructor(arg-list) : base1(arg-list), base2(arg-list), baseN(arg-list);
{
//函数体;
}

上面的形式中，派生类构造函数声明和基类之间是用冒号分隔。在又多个基类的情况下，积累之间以逗号分隔。如下程序所示：

[C++] 纯文本查看 复制代码



#include <iostream>

using namespace std;



class base {

protected:

    int i;

public:

    base(int x) { i = x; cout << "Constructing base\n"; }

    ~base() { cout << "Destructing base\n"; }

};



class derived: public base {

    int j;

public:

    //派生类使用x; y 被传递给 base 的构造函数

    derived(int x, int y): base(y)

    {

        j = x;

        cout << "Constructing derived\n";

    }

    ~derived() { cout << "Destructing derived\n"; }



    void show() { cout << i << " " << j << endl; }

};



int main()

{

    derived ob(3, 4);

    ob.show();

    return 0;

}

运行输出：

$ ./passargtobase
Constructing base
Constructing derived
4 3
Destructing derived
Destructing base

上面程序中，derived 的构造函数里有两个参数 x 和 y 。derived() 只用了参数 x，参数 y 被传递给了基类构造函数 base() 。一般情况下，在派生类的构造函数中，除了声明该类需要的参数外，还必须声明基类需要的参数。

下面程序是派生类从多个基类中继承的示例：

[C++] 纯文本查看 复制代码





#include <iostream>

using namespace std;



class base1 {

protected:

    int i;

public:

    base1(int x) { i = x; cout << "Constructing base\n"; }

    ~base1() { cout << "Destructing base\n"; }

};



class base2 {

protected:

    int k;

public:

    base2(int x) { k = x; cout << "Constructing base2\n"; }

    ~base2() { cout << "Destructing base2\n"; }

};





class derived: public base1, public base2  {

    int j;

public:

    derived(int x, int y, int z): base1(y), base2(z)       //注意 y 和 z 的传递

    {

        j = x;

        cout << "Constructing derived\n";

    }

    ~derived() { cout << "Destructing derived\n"; }



    void show() { cout << i << " " << j << " " << k << endl; }

};



int main()

{

    derived ob(3, 4, 5);

    ob.show();

    return 0;

}

运行输出：

$ ./passargtobase2
Constructing base
Constructing base2
Constructing derived
4 3 5
Destructing derived
Destructing base2
Destructing base

在传递参数给基类构造函数的一个重点是：基类构造函数中的参数是通过派生类的构造函数传递的。基于此，像上面的程序，假如派生类的构造函数 derived() 并不要求带有任何参数，但如果基类的构造函数需要带有一个或多个参数，那么它就必须声明这些参数。如下代码所示：

[C++] 纯文本查看 复制代码





#include <iostream>

using namespace std;



class base1 {

protected:

    int i;

    int j;

public:

    base1(int x, int y) { i = x; j = y; cout << "Constructing base\n"; }

    ~base1() { cout << "Destructing base\n"; }

};



class base2 {

protected:

    int k;

public:

    base2(int x) { k = x; cout << "Constructing base2\n"; }

    ~base2() { cout << "Destructing base2\n"; }

};





class derived: public base1, public base2  {



public:

    derived(int x, int y, int z): base1(x, y), base2(z)

    {

        cout << "Constructing derived\n";      //该构造函数不需要带有参数

    }

    ~derived() { cout << "Destructing derived\n"; }



    void show() { cout << i << " " << j << " " << k << endl; }

};



int main()

{

    derived ob(3, 4, 5);

    ob.show();

    return 0;

}

运行输出：

$ ./passargtobase3
Constructing base
Constructing base2
Constructing derived
3 4 5
Destructing derived
Destructing base2
Destructing base

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