[C++11]继承构造函数

Posted 2021-06-28 Wecccccccc

C++11中提供的继承构造函数可以让派生类直接使用基类的构造函数，而无需自己再写构造函数，尤其是在基类有很多构造函数的情况下，可以极大地简化派生类构造函数的编写。

先来看没有继承构造函数之前的处理方式:
代码如下:

#include <iostream>
#include <string>
using namespace std;

class Base
{
public:
	Base(int i):m_i(i){}
	Base(int i,double j):m_i(i),m_j(j){}
	Base(int i,double j,string k):m_i(i),m_j(j),m_k(k){}

	int m_i;
	double m_j;
	string m_k;
};

class Child :public Base
{
public:
	Child(int i):Base(i){}
	Child(int i,double j):Base(i,j){}
	Child(int i,double j,string k):Base(i,j,k){}

};

int main()
{
	Child c(520, 13.14, "I Love You");
	cout << "int  = " << c.m_i << " double = " << c.m_j << " string  = " << c.m_k << endl;
	return 0;
}

测试结果:

继承构造函数
代码如下:

#include <iostream>
#include <string>
using namespace std;

class Base
{
public:
	Base(int i):m_i(i){}
	Base(int i,double j):m_i(i),m_j(j){}
	Base(int i,double j,string k):m_i(i),m_j(j),m_k(k){}

	int m_i;
	double m_j;
	string m_k;
};

class Child :public Base
{
public:
	using Base::Base;

};

int main()
{
	Child c(520, 13.14, "I Love You");
	cout << "int  = " << c.m_i << " double = " << c.m_j << " string  = " << c.m_k << endl;
	return 0;
}

测试结果:

如果子类中有跟父类一样的同名函数，就会隐藏父类的一样同名的函数。

代码如下:

#include <iostream>
#include <string>
using namespace std;

class Base
{
public:
	Base(){}
	Base(int i):m_i(i){}
	Base(int i,double j):m_i(i),m_j(j){}
	Base(int i,double j,string k):m_i(i),m_j(j),m_k(k){}

	void func(int i)
	{
		cout << "base func(int i)" << endl;
	}

	void func(int i, string str)
	{
		cout << "base func(int i,string str)" << endl;
	}

	int m_i;
	double m_j;
	string m_k;
};

class Child :public Base
{
public:
	using Base::Base;

	void func()
	{
		cout << "child func()" << endl;
	}

};

int main()
{
	Child c;
	c.func();
	//c.func(2);//error
	//c.func(2, "Tom");//error

	return 0;
}

测试结果:

在子类中加入下面这条代码，则可以解除隐藏。

using Base::func;

代码如下:

#include <iostream>
#include <string>
using namespace std;

class Base
{
public:
	Base(){}
	Base(int i):m_i(i){}
	Base(int i,double j):m_i(i),m_j(j){}
	Base(int i,double j,string k):m_i(i),m_j(j),m_k(k){}

	void func(int i)
	{
		cout << "base func(int i)" << endl;
	}

	void func(int i, string str)
	{
		cout << "base func(int i,string str)" << endl;
	}

	int m_i;
	double m_j;
	string m_k;
};

class Child :public Base
{
public:
	using Base::Base;
	using Base::func;
	void func()
	{
		cout << "child func()" << endl;
	}

};

int main()
{
	Child c;
	c.func();
	c.func(2);
	c.func(2, "Tom");

	return 0;
}

测试结果: