程序设计实习

【程序设计实习】第五次上机复盘

02:统计动物数量

描述

代码填空，使得程序能够自动统计当前各种动物的数量

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#include <iostream>
using namespace std;
// 在此处补充你的代码
void print() {
	cout << Animal::number << " animals in the zoo, " << Dog::number << " of them are dogs, " << Cat::number << " of them are cats" << endl;
}

int main() {
	print();
	Dog d1, d2;
	Cat c1;
	print();
	Dog* d3 = new Dog();
	Animal* c2 = new Cat;
	Cat* c3 = new Cat;
	print();
	delete c3;
	delete c2;
	delete d3;
	print();
}

输入 

1

无

输出 

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0 animals in the zoo, 0 of them are dogs, 0 of them are cats
3 animals in the zoo, 2 of them are dogs, 1 of them are cats
6 animals in the zoo, 3 of them are dogs, 3 of them are cats
3 animals in the zoo, 2 of them are dogs, 1 of them are cats
样例输入
None
样例输出
0 animals in the zoo, 0 of them are dogs, 0 of them are cats
3 animals in the zoo, 2 of them are dogs, 1 of them are cats
6 animals in the zoo, 3 of them are dogs, 3 of them are cats
3 animals in the zoo, 2 of them are dogs, 1 of them are cats

Solution

有一点要注意，因为太久没写忘了
就是基类的析构函数要写成虚函数形式，这样，在delete派生类对象的时候才会先执行派生类析构函数，再执行基类的析构函数，消得干净
这里我们都会写 但是通过C++的报错可以得知一个点
print函数中引用的Animal::number这个东西，它是引用的Animal类（划重点）的number成员变量
这个时候就必须写成静态形式，因为如果不写成的话，编译器不知道你调用的是哪个Animal类对象的成员变量
这个也是因为太久没写忘了
下面看代码:

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#include <iostream>
using namespace std;
class Animal{
public:
    static int number;
	Animal(){
		number+=1;
	}
	virtual ~Animal(){
		number--;
	}
};
class Dog:public Animal{
public:
	static int number;
	Dog(){
		number+=1;
	}
	~Dog(){
		number--;
	}
};
class Cat:public Animal{
public:
	static int number;
	Cat(){
		number+=1;
	}
	~Cat(){
		number--;
	}
};
int Animal::number=0;
int Dog::number=0;
int Cat::number=0;
// 在此处补充你的代码
void print() {
	cout << Animal::number << " animals in the zoo, " << Dog::number << " of them are dogs, " << Cat::number << " of them are cats" << endl;
}

int main() {
	print();
	Dog d1, d2;
	Cat c1;
	print();
	Dog* d3 = new Dog();
	Animal* c2 = new Cat;
	Cat* c3 = new Cat;
	print();
	delete c3;
	delete c2;
	delete d3;
	system("pause");
	print();
}

04:多态

描述 

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#include <iostream>
using namespace std;

class Base {
public:
   virtual Base& fun() { cout << "base fun" << endl; return *this; }
   virtual Base& foo() { cout << "base foo" << endl; return *this; }
};

class Derived: public Base {
public:
   Base& fun() { cout << "derived fun" << endl; return *this; }
   Base& foo() { cout << "derived foo" << endl; return *this; }
};

Base& foo();
Base& fun();
// 在此处补充你的代码
int main() {
   foo().fun().foo();
   fun().foo().fun();
   return 0;
}

输入 

1

-

输出 

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derived foo
derived fun
derived foo
base fun
base foo
base fun

样例输入 

1

-

样例输出 

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derived foo
derived fun
derived foo
base fun
base foo
base fun

Solution

这道题花的时间最长
估计是因为函数的写了不调用形式没怎么见到（以及python写太久了，C++有所淡忘）
就是那个Base& foo();
这个表示这是个函数 但是啥都没声明
所以，题目就是需要我们声明一下两个函数是什么
然后呢？我们看一下这个函数的返回值类型
当时就是因为没注意到这个，所以好长时间没做出来
我们以第二个函数fun为例
它声明一个Base类对象b 首先，为什么要静态？
因为这个静态对象只能创建一次，并且在函数结束后不会被销毁
然后return b.fun();
这句的意思是执行一次b的fun（Base类）函数，再返回一个b对象的拷贝
而Derived类和Base类中的foo/fun成员函数同理
所以才会输出三次一样的
下面看代码:

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#include <iostream>
using namespace std;

class Base {
public:
   virtual Base& fun() { cout << "base fun" << endl; return *this; }
   virtual Base& foo() { cout << "base foo" << endl; return *this; }
};

class Derived: public Base {
public:
   Base& fun() { cout << "derived fun" << endl; return *this; }
   Base& foo() { cout << "derived foo" << endl; return *this; }
};

Base& foo();
Base& fun();
Base& foo(){
   static Derived d;
   return d.foo();
}
Base& fun(){
   static Base b;
   return b.fun();
}
// 在此处补充你的代码
int main() {
   foo().fun().foo();
   fun().foo().fun();
   system("pause");
   return 0;
}

06:变来变去的数

描述

程序填空输出指定结果。（行末输出多余的空格不会影响结果的正确性）

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#include <iostream>
using namespace std;

class myobject {
public:
// 在此处补充你的代码
};

class producer : public myobject {
public:
	virtual void work() {
		counter = counter + 5;
		print_avaliable();
	}
};

int myobject::counter = 0;

int main(){
	producer *pro = new producer();
	myobject *con = new myobject();
	pro->work(); pro->work(); cout << endl;
	con->work(); con->work(); con->work(); cout << endl;
	pro->work(); cout << endl;
	con->work(); con->work(); cout << endl;
}

输入 

1

无

输出 

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5 10
6 2 2
7
3 3

样例输入 

1

无

样例输出 

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5 10
6 2 2
7
3 3

Solution

这里其实没有啥要看清的
就是抖机灵 在A的函数里写上counter大于等于4就减一这种解法

  • 笔者注:此题后得知可不用特判解出，本解法仅供参考

下面看代码：

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#include <iostream>
using namespace std;

class myobject {
public:
    static int counter;
    void work(){
        if(counter>=4) counter-=4;
        cout<<counter<<' ';
    }
    void print_avaliable(){
        cout<<counter<<' ';
    }
// 在此处补充你的代码
};

class producer : public myobject {
public:
	virtual void work() {
		counter = counter + 5;
		print_avaliable();
	}
};

int myobject::counter = 0;

int main(){
	producer *pro = new producer();
	myobject *con = new myobject();
	pro->work(); pro->work(); cout << endl;
	con->work(); con->work(); con->work(); cout << endl;
	pro->work(); cout << endl;
	con->work(); con->work(); cout << endl;
    system("pause");
    return 0;
}

07:MyClass

描述

补充下列代码，使得程序的输出为： A:3 A:15 B:5 3 15 5

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#include <iostream>
using namespace std;
class CMyClassA {
	int val;
public:
	CMyClassA(int);
	void virtual print();
};
CMyClassA::CMyClassA(int arg) {
	val = arg;
	printf("A:%d\n", val);
}
void CMyClassA::print() {
	printf("%d\n", val);
	return;
}
// 在此处补充你的代码
int main(int argc, char** argv) {
	CMyClassA a(3), *ptr;
	CMyClassB b(5);
	ptr = &a; ptr->print();
	a = b;
	a.print();
	ptr = &b; ptr->print();
	return 0;
}

输入 

1

无

输出 

1

见样例

样例输入 

1

None

样例输出 

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A:3
A:15
B:5
3
15
5

Solution

别的不说，就说一个
这里的派生类的构造函数如何调用基类构造函数？
做的时候忘记了
下面看代码:

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#include <iostream>
using namespace std;
class CMyClassA {
	int val;
public:
	CMyClassA(int);
	void virtual print();
};
CMyClassA::CMyClassA(int arg) {
	val = arg;
	printf("A:%d\n", val);
}
void CMyClassA::print() {
	printf("%d\n", val);
	return;
}
class CMyClassB:public CMyClassA{
public:
    int val;
    CMyClassB(int x):CMyClassA(3*x){
        val=x;
        printf("B:%d\n",val);
    }
    void print(){
        printf("%d\n",val);
    }
};
// 在此处补充你的代码
int main(int argc, char** argv) {
	CMyClassA a(3), *ptr;
	CMyClassB b(5);
	ptr = &a; ptr->print();
	a = b;
	a.print();
	ptr = &b; ptr->print();
    system("pause");
	return 0;
}
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