1、面向對象方法與C+題庫及答案、填空題1 .若char *string= test ”;則如果要輸出指針值，正確的語句是 。2 .在重載“口 ”運算符時，必須使用 函數重載。3 .當用public繼承時，基類的public成員成為派生類的 成員，基類的protected 成員成為派生類的 成員。4 .可以賦給指針的唯一整數是 。5 .在重載“=”運算符時，必須使用 函數重載。6 .以下程序：int c=10;cout.flags(ios:hex|ios:showbase);cout XXXX;XXXX必須是一個變量名，而不能是一個任意表達式。()2 .以下聲明將不會導致編譯錯誤。()enum G

2、radeType A, B, C, D, E, F;3 .在C+理，派生類的構造函數先于其基類的構造函數執行。()4 . C+中的所有函數都是傳值調用。()5 .在類time中聲明如下的函數原型：void time();()6 .以下程序：class CA public:virtual void dis( )=0; ;class CB:public CA public:void set(int m) k=m; protected:int k;；void main() CB b; 是正確的。（）7 .函數模板能夠定義一個在不同數據類型基礎上完成同一任務的函數。（）8 .以下語句：char *st

3、ring= test ;delete string;是正確的 。（）9 .一個有指針數據成員的類必須要包含的成員函數有：初始化構造函數，析構函數，拷貝構造函數和賦值運算符函數。（）10 .抽象基類中的所有虛函數都要聲明為純虛函數。（）11 .以下語句：char *string=new char10;strcpy（string, test ）;/ delete string;是正確的 。（）12 .枚舉類型變量的值可以直接進行輸入、輸出。（） 三、讀程題1. .下面程序在執行過程中，若順序輸入十個整數:1 2 3 4 5 6 7 8 9 10,其結果是什么。#include class stac

4、k;class nodeint data;node *prev;public:node(int d,node *n) data=d; prev=n;friend class stack;；class stacknode *top;public:stack( ) top=0;void push(int i)node *n=new node(i,top); top=n; int pop( )node *t=top;if(top)top=top-prev; int c=t-data;delete t; return c;return 0; ;void main( )int c; stack s;fo

5、r(int i=0;ic; s.push(c); for(i=0;i10;i+)cout s.pop( ) coutendl; 2.#include #include class Increment public:Increment(int=0,int=1);void addIncrement() count+=increment;void print( ) const;private:int count;const int increment; ;Increment:Increment(int c,int i):increment(i) count=c; void Increment:pri

6、nt( ) constcoutcount=setw(8)countendl;coutincrement=incrementendl; void main( )Increment object(65,7);coutbefore:endl;object.print();for(int j=0;j3;j+)object.addIncrement();coutafter:j+1endl;object.print( ); 3.#include class pointstatic int count ;float xcoord , ycoord ;public:point(float x=0 float

7、y=0 ) xcoord=x ;ycoord=y ;count+ ; static int getcount() return count ;point() count- ; ;int point:count=0;void main( )coutpoint:getcount( ); point *p ,a(32.98,-4.71) ,b ,c; coutpoint:getcount( ); coutb.getcount( );p=new point100;coutpoint:getcount( );delete p;coutpoint:getcount( )endl; 4.#include c

8、lass Teststatic int count;public:Test( ) +count;Test( ) -count; static int getCount( ) return count; ;int Test:count=0;void main( )coutTest:getCount();Test t,tab5,*p;couttTest:getCount();p=new Test10;couttTest:getCount();delete p;couttTest:getCount( )endl; 5.#include #include template class vector T

9、 *v; int sz;public:vector(int s=100) v=new Tsz=s; vector(const vector&vv)sz=vv.sz; v=new Tsz;vector&operator=(const vector&vv)if(&vv!=this)delete v;sz=vv.sz;v=new Tsz;for(int i=0;isz;i+) vi=vv.vi; return *this;vector() delete v; T & operator (int i) if(i=sz) cerrErrorn;exit(1); return vi; ;void main

10、( )vector v(10),u(v);for(int i=0;i10;i+)vi=3*i+1;u=v;for( i=0;i=10;i+) coutui )和插入符( )以支持I/O 操作。在 main函數中定義學生對象數組，對于學生對象數組從磁盤文件 “student.txt ”進行輸入，最后以學號n為參數在數組中查找學號為n的學生，并顯示該生的全部信息。#include #include #include class student 2； char name20,sex; int age;float score;public:student( ) student(int nu,char

11、* na,char se,int ag,3 sc) _num=nu; strcpy(name,na);sex=se; age=ag; score=sc; int getn( )return num;friend istream&operator(istream& s,student& st)st.sexst.agest.score;return 4; friend ostream&operator(5 s,const student& st) sst.num 6 st.sex st.age st.scores_9_;_int n,flag=1;coutEnter n

12、:;10;for(int j=0;ji & flag;j+)if(sj.getn( )=n) flag=0;cout找到了 :sj;if(flag=1) cout沒找到。n;2 .請完成以下復數類的設計，復數類對象的創建及使用。#include #include #include class complexdouble rpart,ipart;double abs( )const;double norm( )const;public:complex(double r=0,double i=0)2; ipart=i; complex operator-();friend complex oper

13、ator+(const complex&,const complex&);friend complex operator-(const complex&,const complex&);friend complex operator*(const complex&,const complex&);friend complex operator/(const complex&,const complex&);friend 3 operator(const complex&,const complex&);friend int operator=(const complex&,const comp

14、lex&);friend int operator (const complex&,const complex&);friend int operator(istream& si,complex& c)sic.rpartc.ipart;return si;friend ostream& operator( ostream&so,const complex&c) so(c.rpart,c.ipart)n;4; ;double complex:abs( )constreturn sqrt(rpart*rpart +5);_double complex:norm( )constreturn (rpa

15、rt*rpart+ipart*ipart);complex complex: operator-( )return complex(-rpart,-ipart);complex operator+(const complex&c1,const complex&c2)return complex(c1.rpart+c2.rpart),c1.ipart+c2.ipart);complex operator-(const complex&c1,const complex&c2)return complex(6 ,c1.ipart-c2.ipart);complex operator*(const c

16、omplex&c1,const complex&c2)return complex(c1.rpart*c2.rpart-c1.ipart*c2.ipart,c1.rpart*c2.ipart+c1.ipart*c2.rpart); complex operator/(const complex&c1,const complex&c2)complex res;double d=c2.norm();if(d!=0)res.rpart=(c1.rpart*c2.rpart+c1.ipart*c2.ipart)/d;res.ipart=(c1.ipart*c2.rpart-c1.rpart*c2.ip

17、art)/d; else cerr(const complex&c1,const complex&c2)return c1.abs( )c2.abs( );int operator=(const complex&c1,const complex&c2)return 8; int operator(const complex&c1,const complex&c2)return c1.abs( )c2.abs( ); int operator=(const complex&c1,const complex&c2)return c1.abs( )c3; coutc3;c3=-c1*2+c2/c3*

18、c1-6;coutc2) coutc1;else coutc1c2c3;coutc1c2c3; 3 .有一家醫院的門診記錄如下所示，已知該記錄存放在一個patient.dat ”文件中，請編寫一個程序，幫助醫生計算一下每一位病人平均血壓。病人的編號病人血壓的測量次數病人血壓各次測量的結果10015100 120 90 110 10010022100 1201003390 120 1301004480 70 90 10010053120 135 1104 .請完成以下集合類的設計，集合類對象的創建及使用。#include const int maxcard=20;class setint ele

19、msmaxcard,card;public:set( ) card=1;friendbool operator&(int,set);friendbool operator=(set,set);friend 2operator!=(set,set);friendsetoperator+(set,set);friendsetoperator+(set,int);friendsetoperator-(set,int);3 setoperator*(set,set);friendbool operator(set,set);friendbool operator=(set,set);friendost

20、ream& operator(ostream&out,constset&s) outendl;for(int i=0;is.card;i+)outs.elemsi;return 4; _;bool operator&(int e ,set s)for(int i=0;is.card;i+)if(s.elemsi=e)return true;return 5;bool operator=(set s1,set s2)if(s1.card!=s2.card) return false;for(int i=0;is1.card;6)_if(!(s1.elemsi&s2) return false;r

21、eturn true;bool operator!=(set s1,set s2) return !(s1=s2)?true:false;set operator+(set s,int e) set res=s;if(s.cardmaxcard)if(!(e&s)res.elemsres.card+=e;return res;set operator+(set s1,set s2) set res=s1;for(int i=0;is2.card;i+)res=res+s2.elemsi;return res;set operator-(set s,int e) set res=s;if(!(e

22、&s)return res;for(int i=0;is.card;i+)if(s.elemsi=e)for(;is.card-1;i+) res.elemsi=res.elemsi+1;7;return res;set operator*(set s1,set s2)8;for(int i=0;is1.card;i+)for(int j=0;js2.card;j+)if(s1.elemsi=s2.elemsj) res.elemsres.card+=9; break; return res;bool operators2.card)return false;for(int i=0;is1.c

23、ard;i+)if(!(s1.elemsi&s2)return false;return true;bool operator(set s1,set s2)return s1.cards2.card&s1=s2?true:false;void main( ) set 10 ;for(int i=0;i100;i+) s1=s1+i;couts1;if(s1!=s2) coutnTruen; else coutnFalsen;s2=s1; couts2;if(s1=s2) coutnTruen; else coutnFalsen; for(i=0;i20;i+) s1=s1-i;couts1;s

24、3=s1*s2; couts3;if(s1s2) coutnTruen; else coutnFalsen;if(s1+s2=s3)coutnTruen; else coutnFalsen; 5.完成以下程序。#include #include double function(double x)return 4/(1+x*x); class inte_algo protected:double a,b,n,h,sum;public:inte_algo(double left,double right,double steps)a=1,b=right,n=steps,h=(b-a)/n,sum=

25、0;virtual void integrate( )=2; ; _ class rectangle:3 public:rectangle(double left,double right,double steps): inte_algo(left,right,steps) 4; ;void rectangle:integrate( ) double a1=a;for(int i=0;in;i+)sum+=function(a1),a1+=h;sum*=h;coutSum=sumendl; class ladder: public inte_algo public:ladder(double

26、left,double right,double steps): inte algo(left,right,steps) void integrate( ); ;void ladder:integrate( ) double a1=a;sum=(function(a)+function(b)/2;for(int i=1;in;i+)a1+=h,sum+=function(a1);sum*=h;coutSum=sumendl; class 5 : public inte_algo public:simpson(double left,double right,double steps):6 vo

27、id integrate( ); ;void simpson:integrate( ) sum=function(a)+function(b);double s=1,a1=a;for(int i=1;in;i+)a1+=h, sum+=(3+s)*function(a1),s=-s;sum*=h/=3;_7_； void main( ) rectangle r(0,1,10); ladder l(0,1,10);simpson 8(0,1,10); inte_algo 9=&r;coutintegrate( ); p=&l;p-integrate( ); p=&s;10; 第12頁共14頁面向對象方法與C+作業參