C++的探索路13继承与派生之练习篇(需重新学习)

Posted 2022-12-10 Guerrouj

由程序结果填空

输出：4，6

填空：

class A 
	int val;
public:
	A(int n) 
		val = n;
	
	int GetVal() 
		return val;
	
;
class B :public A 
private:
	int val;
public:
	B(int n) :____ 

	
	int GetVal() 
		return val;
	
;
int main() 
	B b1(2);
	cout << b1.GetVal() << "," << b1.A::GetVal() << endl;
	return 0;

输出4，6则说明

b1.GetVal()输出4，而b1.A::GetVal()输出6

A类包含成员变量val，以及返回val值的成员函数GetVal()

B类继承A类的衣钵，并同样拥有GetVal函数返回val。

先看一下缺失部分：B类对A的初始化。

再看下输出：4，6；恰好为2的两倍与三倍。

因此填空可以填为：

	B(int n) : A(3*n)
		val = 2 * n;
	

写出程序的输出结果

class Base 
public:
	int val;
	Base() 
		cout << "Base Constructor" << endl;
	
	~Base() 
		cout << "Base Destructor" << endl;
	
;
class Base1 :virtual public Base ;
class Base2 :virtual public Base ;
class Derived:public Base1,public Base2;
int main() 
	Derived d;
	return 0;

由于采用了虚继承，因此只会构造与析构一次。

输出结果应当为

Base Constructor

Base Destructor

重新编写MyString类

这是一道复合题

第一问来自于运算符重载：

编写一个MyString类，输出结果为:

1.abcd-efgh-abcd-

2.abcd-

3.

4.abcd-efgh-

5.efgh-

6.c

7.abcd-

8.ijAl-

9.ijAl-mnop

10.qrst-abcd

11.abcd-qrst-abcd-uvw xyz

about

big

me

take

abcd

qrst-abcd-

第二问是

MyString类必须从string类派生而来。

提示1，如果MyString由string替换，除了最后两行无法编译通过，其他都可以。也就是扩充体现在最后两条语句上面。

提示2，string类有一个成员函数string substr(int start,int length)，能够求从start开始，长度为length的子串

不要问我为什么这么长，我也很绝望啊~~

程序如下

#include<cstring>
#include<cstdlib>
using namespace std;
int CompareString(const void*e1, const void*e2) 
	MyString*s1 = (MyString*)e1;
	MyString*s2 = (MyString*)e2;
	if (*s1 < *s2) return -1;
	else if (*s1 == *s2)return 0;
	else if (*s1 > *s2)return 1;

int main() 
	MyString s1("abcd-"), s2,
		s3("efgh-"), s4(s1);
	MyString SArray[4] = 
		"big","me","about","take"
	;
	cout << "1." << s1 << s2 << s3 << s4 << endl;
	s4 = s3; s3 = s1 + s3;
	cout << "2. " << s1 << endl;
	cout << "3. " << s2 << endl;
	cout << "4. " << s3 << endl;
	cout << "5. " << s4 << endl;
	cout << "6. " << s1[2] << endl;
	s2 = s1;
	s1 = "ijkl-";
	s1[2] = 'A';
	cout << "7. " << s2 << endl;
	cout << "8. " << s1 << endl;
	s1 += "mnop";
	cout << "9. " << s1 << endl;
	s4 = "qrst-" + s2;
	cout << "10. " << s4 << endl;
	s1 = s2 + s4 + "uvw" + "xyz";
	cout << "11. " << s1 << endl;
	qsort(SArray, 4, sizeof(MyString), CompareString);
	for (int i = 0; i < 4; ++i)
		cout << SArray[i] << endl;
	cout << s1(0, 4) << endl;//输出s1从下标0开始长度为4的子串
	cout << s1(5, 10) << endl;
	return 0;

emmm。。。智商有点不够用了，直接贴答案，后续将对这部分进行再次练习

原文链接  程序设计实习MOOC / 继承和派生——编程作业 第五周程序填空题1

参考coding

#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
class MyString : public string

public:
	MyString() ;
	//1.0继承类继承父类所有的成员变量和成员函数，但不继承构造函数和析构函数 
	//1.1继承类的无参构造函数，会隐式调用父类的无参构造函数 
	MyString(const char * st) :string(st) ;
	//1.2继承类的有参构造函数，如果父类也有有参构造函数，则必须显示调用它 
	//2.0这里的参数根据reference有两种选择，此处必须用const char*,"xxx"的类型是const char* 
	MyString(const MyString& s) :string(s) 
	//1.3继承类的复制构造函数必须要显示的调用父类的复制构造函数，不然就会默认调用父类的无参构造函数 
	MyString operator +(MyString & op2)
	
		string s1 = *this;
		string s2 = op2;
		string s = s1 + s2;
		return *new MyString(s.c_str());
	
	MyString & operator +(const char * cs2)
	
		string str1 = *this;
		string s = str1 + cs2;
		return *new MyString(s.c_str());
	

	MyString & operator()(int s, int l)
	
		string str = substr(s, l);
		return *new MyString(str.c_str());
	
;

MyString operator+(const char * op1, MyString & op2)

	string st2 = op2;
	string s = op1 + st2;
	return *new MyString(s.c_str());


int CompareString(const void * e1, const void * e2)

	MyString * s1 = (MyString *)e1;
	MyString * s2 = (MyString *)e2;
	if (*s1 < *s2) return -1;
	else if (*s1 == *s2) return 0;
	else if (*s1 > *s2) return 1;

int main()

	MyString s1("abcd-"), s2, s3("efgh-");
	MyString s4(s1);
	MyString SArray[4] =  "big","me","about","take" ;
	//这里等号右边的赋值操作相当于调用了MyString的转换构造函数，其实就是单一非const classname&参数的构造函数可以直接接受参数类型的变量
	cout << "1. " << s1 << s2 << s3 << s4 << endl;
	s4 = s3;
	//3.0 operator=可以直接用string类里面的 
	s3 = s1 + s3;
	s1 + s3;
	cout << "2. " << s1 << endl;
	cout << "3. " << s2 << endl;
	cout << "4. " << s3 << endl;
	cout << "5. " << s4 << endl;
	cout << "6. " << s1[2] << endl;
	s2 = s1;
	s1 = "ijkl-";
	s1[2] = 'A';
	cout << "7. " << s2 << endl;
	cout << "8. " << s1 << endl;
	s1 += "mnop";
	cout << "9. " << s1 << endl;
	s4 = "qrst-" + s2;
	cout << "10. " << s4 << endl;
	s1 = s2 + s4 + " uvw " + "xyz";
	cout << "11. " << s1 << endl;
	qsort(SArray, 4, sizeof(MyString), CompareString);
	for (int i = 0; i < 4; ++i)
		cout << SArray[i] << endl;
	cout << s1(0, 4) << endl;
	cout << s1(5, 10) << endl;
	return 0;