作业2.5

第四章  堆与拷贝构造函数

 一 、程序阅读题

1、给出下面程序输出结果。

#include

class example

{int a;

public:

example(int b=5){a=b++;}

void print(){a=a+1;cout <

void print()const

{cout<

};

void main()

{example x;

const example y(2);

x.print();

y.print();

}

答：6，3

2、运行程序，写出程序执行的结果。

#include

class Location

{   public:

int X,Y;

void init(int initX,int initY);

int GetX();

int GetY();

};

void Location::init (int initX,int initY)

{X=initX;

Y=initY;

}

int Location::GetX()

{return X;

}

int Location::GetY()

{return Y;

}

void display(Location& rL)

{cout<＼n';

}

void main()

{

Location A［5］={{5,5},{3,3},{1,1},{2,2},{4,4}};

Location *rA=A;

A［3］.init(7,3);

rA->init(7,8);

for (int i=0;i<5;i++)

display(*(rA++));

}

答：

7 8

3 3

1 1

7 3

4 4

3. 给出下面程序输出结果。

#include

int a[8］={1,2,3,4,5,6,7};

void fun(int *pa,int n);

void main()

{int m=8;

fun(a,m);

cout<［7］<

}

void fun(int *pa,int n)

{for (int i=0;i

*(pa+7)+=*(pa+i);

}

答：28

4. 给出下面程序输出结果。

#include

class A

{

int *a;

public:

A(int x=0):a(new int(x)){}

~A() {delete a;}

int getA() {return *a;}

void setA(int x) {*a=x;}

};

void main()

{

A x1,x2(3);

A *p=&x2;

(*p).setA(x2.getA()+5);

x1.setA(10+x1.getA());

cout<

}

答：108

5. 阅读下面的程序，写出运行结果：

#include < iostream.> using namespace std; class Samp { public:     void Set_i_j(int a, int b){i=a,j=b;}     ~Samp()     {         cout <<"Destroying.." << i <     }     int GetMulti () { return i * j; } protected: int i; int j; }; int main () { Samp * p; p = new Samp[l0]; if(!p) { cout << "Allocation error \ n"; return; } for(int j =0; j     p[j]. Set_i_j (j, j); for(int k=0; k     cout <<"Multi[" < delete [ ] p; return 0; }

 答：

Multi[0] is: 0

Multi[1] is: 1

Multi[2] is: 4

Multi[3] is: 9

Multi[4] is: 16

Multi[5] is: 25

Multi[6] is: 36

Multi[7] is: 49

Multi[8] is: 64

Multi[9] is: 81

Destroying..9

Destroying..8

Destroying..7

Destroying..6

Destroying..5

Destroying..4

Destroying..3

Destroying..2

Destroying..1

Destroying..0

6. 写出下面程序的运行结果，请用增加拷贝构造函数的方法避免存在的问题。

#include < iostream> using namespace std; class Vector { public:     Vector (int s = 100);     int& Elem(int ndx);     void Display();     void Set ();     ~Vector (); protected: int size; int* buffer; } Vector::Vector (int s) { buffer = new int [size = s]; for(int i = O; i     buffer [i] = i* i; } int& Vector:: Elem(int ndx) {     if(ndx< 0 || ndx> = size)     {         cout << "error in index" <         exit (1);     }     return buffer [ndx]; } void Vector::Display () {     for(int j =0; j< size; j ++)     cout << buffer[j] < } void Vector:: Set () {     for(int j =0; j         buffer[j] = j + 1; } Vector:: ~ Vector() {     delete [] buffer; } int main() {     Vector a(10);     Vector b(a);     a. Set ();     b. Display ();   return 0; }

 答：

结果：0 1 4 9 16 25 36 49 64 81

拷贝构造函数：

Vector::Vector(const Vector& other) ：size(other.size),buffer(other,buffer) {}

7.读下面的程序与运行结果，添上一个拷贝构造函数来完善整个程序。

   

#include < iostream> using namespace std; class CAT { public:     CAT();     CAT(const CAT&);     ~CAT();     int GetAge() const (return * itsAge;)     void SetAge(int age) { * itsAge = age; } protected: int * itsAge; }; CAT::CAT () { itsAge = new int; *itsAge = 5; } CAT::~CAT () {     delete itsAge;     itsAge = 0; } void main() {     CAT frisky;     cout << "frisky's age:" << frisky. GetAge() <     cout <<"Setting frisky to 6... \ n";     frisky. SetAge ( 6 );     cout << "Creating boots from frisky \ n";     CAT boots(frisky);     cout <<"frisky's age:" << frisky. GetAge() <     cout << "boots'age:" << boons. GetAge () <     cout << "setting frisk,, to 7 .... n";     frisky. SetAge (7);     cout <<"frisky"s age:" << frisky. GetAge() <     cout <<"boots' age:" << boots. GetAge() < }

 

运行结果为：

    frisky's age:5

    Setting frisky to 6...

    Creating boots from frisky

    frisky's age:6

    boots' age:6

    Setting frisky to 7...

    frisky's age:7

    boots' age:6

 拷贝构造函数：

CAT::CAT(const CAT &other):itsAge(other.itsAge){}