程序設計教程機械工業(yè)出社課后習題答案 操作符重載

1、第7章 操作符重載1、 為什么要對操作符進行重載？是否所有的操作符都可以重載？答：通過對C+操作符進行重載，我們可以實現(xiàn)用C+的操作符按照通常的習慣來對某些類（特別是一些數(shù)學類）的對象進行操作，從而使得程序更容易理解。除此之外，操作符重載機制也提高了C+語言的靈活性和可擴充性，它使得C+操作符除了能對基本數(shù)據(jù)類型和構造數(shù)據(jù)類型進行操作外，也能用它們來對類的對象進行操作。 不是所有的操作符都可以重載，因為“.”， “.*”， “:”， “?:”，sizeof這五個操作符不能重載。2、 操作符重載的形式有哪兩種形式？這兩種形式有什么區(qū)別？答：一種就是作為成員函數(shù)重載操作符；另一種就是作為全局（友元

2、）函數(shù)重載操作符。當操作符作為類的非靜態(tài)成員函數(shù)來重載時，由于成員函數(shù)已經(jīng)有一個隱藏的參數(shù)this，因此對于雙目操作符重載函數(shù)只需要提供一個參數(shù)，對于單目操作符重載函數(shù)則不需提供參數(shù)。當操作符作為全局函數(shù)來重載時，操作符重載函數(shù)的參數(shù)類型至少有一個為類、結構、枚舉或它們的引用類型。而且如果要訪問參數(shù)類的私有成員，還需要把該函數(shù)說明成相應類的友元。對于雙目操作符重載函數(shù)需要兩個參數(shù)，對于單目操作符重載函數(shù)則需要給出一個參數(shù)。操作符=、 ( )、 以及->不能作為全局函數(shù)來重載。另外，作為類成員函數(shù)來重載時，操作符的第一個操作數(shù)必須是類的對象，全局函數(shù)重載則否。3、 定義一個時間類Time，

3、通過操作符重載實現(xiàn)：時間的比較（=、!=、>、>=、<、<=）、時間增加/減少若干秒（+=、-=）、時間增加/減少一秒（+、-）以及兩個時間相差的秒數(shù)（-）。解：class Timeprivate :int hour, minute, second;public :Time()hour=minute=second=0;Time(int h)hour=h;minute=second=0;Time(int h, int m)hour=h;minute=m;second=0;Time(int h, int m, int s)hour=h;minute=m;second=s;T

4、ime(const Time &t)hour=t.hour;minute=t.minute;second=t.second;bool operator =(Time &t)if (hour=t.hour && minute=t.minute && second=t.second)return true;return false;bool operator !=(Time &t)return !(*this = t);bool operator > (Time &t)if (hour>t.hour)return true

5、;else if (hour=t.hour && minute>t.minute)return true;else if (hour=t.hour && minute=t.minute && second>t.second)return true;elsereturn false;bool operator >=(Time &t)return *this > t | *this = t;bool operator <(Time &t) return !(*this >= t);bool oper

6、ator <=(Time &t)return !(*this > t);Time &operator +=(int s)second+=s;while (second>=60)second-=60;minute+;while (minute>=60)minute-=60;hour+;while (hour>=24)hour-=24;return *this;Time &operator -=(int s)second-=s;while (second<0)second+=60;minute-;while (minute<0)mi

7、nute+=60;hour-;while (hour<0)hour+=24;return *this;Time &operator +() /對Time t，操作為：+t*this += 1;return *this;Time operator +(int) /對Time t，操作為：t+Time t=*this;*this += 1;return t;Time &operator -() *this -= 1;return *this;Time operator -(int) Time t=*this;*this -= 1;return t;int operator -

8、(Time &t)/把時間直接換算成秒數(shù)計算int sec1=hour*3600+minute*60+second;int sec2=t.hour*3600+t.minute*60+t.second;return sec2-sec1;4、 利用操作符重載給出一個完整的復數(shù)類的定義。解：class Complexprivate :double real, imag;public :Complex()real = imag = 0;Complex(double r)real = r;imag = 0;Complex(double r, double i)real = r;imag = i;

9、Complex(const Complex &c)real=c.real;imag=c.imag; double modulus() const return real*real+imag*imag; Complex operator -() constComplex temp;temp .real = -real;temp.imag = -imag;return temp; friend bool operator =(const Complex &c1,const Complex &c2); friend bool operator !=(const Complex

10、 &c1,const Complex &c2); friend bool operator > (const Complex &c1,const Complex &c2); friend bool operator >=(const Complex &c1,const Complex &c2); friend bool operator < (const Complex &c1,const Complex &c2); friend bool operator <=(const Complex &c1

11、,const Complex &c2);friend Complex operator+(const Complex &c1,const Complex &c2);friend Complex operator-(const Complex &c1,const Complex &c2);friend Complex operator*(const Complex &c1,const Complex &c2);friend Complex operator/(const Complex &c1,const Complex &

12、c2);bool operator =(const Complex &c1,const Complex &c2)return (c1.real = c2.real) && (c1.imag = c2.imag);bool operator !=(const Complex &c1,const Complex &c2)return !(c1 = c2);bool operator >(const Complex &c1,const Complex &c2)return c1.modulus() > c2.modulus(

13、);bool operator >=(const Complex &c1,const Complex &c2)return c1.modulus() >= c2.modulus();bool operator <(const Complex &c1,const Complex &c2)return c1.modulus() < c2.modulus();bool operator <=(const Complex &c1,const Complex &c2)return c1.modulus() <= c2.m

14、odulus();Complex operator+(const Complex &c1,const Complex &c2)Complex temp;temp.real=c1.real+c2.real;temp.imag=c1.imag+c2.imag;return temp;Complex operator-(const Complex &c1,const Complex &c2)Complex temp;temp.real=c1.real-c2.real;temp.imag=c1.imag-c2.imag;return temp;Complex opera

15、tor*(const Complex &c1,const Complex &c2)Complex temp;temp.real=c1.real*c2.real-c1.imag*c2.imag;temp.imag=c1.real*c2.imag+c1.imag*c2.real;return temp;Complex operator/(const Complex &c1,const Complex &c2)double d=c2.modulus(); if (d != 0) Complex temp; temp.real=(c1.real*c2.real+c1.i

16、mag*c2.imag)/d; temp.imag=(c1.imag*c2.real-c1.real*c2.imag)/d; return temp; else cout << "Error in operation / of Complex" << endl; exit(-1); 5、 定義一個多項式類Polynomial，其實例為多項式：a0+a1x+a2x2+.+anxn，該類具有如下的接口：class Polynomial .public:Polynomial();Polynomial(double coefs, int exps, int

17、size); /系數(shù)數(shù)組、指數(shù)數(shù)組和項數(shù)Polynomial(const Polynomial&); Polynomial();Polynomial& operator=(const Polynomial&);int degree() const; /最高冪指數(shù)double evaluate(double x) const; /計算多項式的值bool operator=(const Polynomial&) const;bool operator!=(const Polynomial&) const;Polynomial operator+(const

18、Polynomial&) const;Polynomial operator-(const Polynomial&) const;Polynomial operator*(const Polynomial&) const;Polynomial& operator+=(const Polynomial&);Polynomial& operator-=(const Polynomial&);Polynomial& operator*=(const Polynomial&);解：class Polynomialdouble *p

19、coefs;int *pexps;int num_of_items; int add(const Polynomial &p, double *coefs, int *exps) const; int subtract(const Polynomial &p, double *coefs, int *exps) const;public :Polynomial();Polynomial(double coefs, int exps, int size);Polynomial(const Polynomial &p);Polynomial();Polynomial &am

20、p;operator=(const Polynomial &p);int degree() const;double evaluate(double x) const;bool operator=(const Polynomial &p) const;bool operator!=(const Polynomial &p) const;Polynomial operator+(const Polynomial &p) const;Polynomial operator-(const Polynomial &p) const;Polynomial oper

21、ator*(const Polynomial &p) const;Polynomial &operator+=(const Polynomial &p);Polynomial &operator-=(const Polynomial &p);Polynomial &operator*=(const Polynomial &p);int Polynomial:add(const Polynomial &p, double *coefs, int *exps) const int count=0,i=0,j=0; while (i&l

22、t;num_of_items && j<p.num_of_items) if (pexpsi = p.pexpsj) if (pcoefsi != -p.pcoefsj) coefscount = pcoefsi + p.pcoefsj; expscount = pexpsi; count+; i+; j+; else if (pexpsi < p.pexpsj) coefscount = pcoefsi; expscount = pexpsi; count+; i+; else coefscount = p.pcoefsj; expscount = p.pexps

23、j; count+; j+; if (i<num_of_items) while (i<num_of_items) coefscount = pcoefsi; expscount = pexpsi; count+; i+; else while (j<p.num_of_items) coefscount = p.pcoefsj; expscount = p.pexpsj; count+; j+; return count;int Polynomial:subtract(const Polynomial &p, double *coefs, int *exps) con

24、st int count=0,i=0,j=0; while (i<num_of_items && j<p.num_of_items) if (pexpsi = p.pexpsj) if (pcoefsi != p.pcoefsj) coefscount = pcoefsi - p.pcoefsj; expscount = pexpsi; count+; i+; j+; else if (pexpsi < p.pexpsj) coefscount = pcoefsi; expscount = pexpsi; count+; i+; else coefscount

25、 = -p.pcoefsj; expscount = p.pexpsj; count+; j+; if (i<num_of_items) while (i<num_of_items) coefscount = pcoefsi; expscount = pexpsi; count+; i+; else while (j<p.num_of_items) coefscount = -p.pcoefsj; expscount = p.pexpsj; count+; j+; return count;Polynomial:Polynomial()pcoefs=NULL;pexps=NU

26、LL;num_of_items=0;Polynomial:Polynomial(double coefs, int exps, int size)num_of_items=size;pcoefs=new doublenum_of_items;pexps=new intnum_of_items; int i;for (i=0; i<num_of_items; i+)pcoefsi=coefsi;pexpsi=expsi; /按指數(shù)排序（冒泡排序） for (i=num_of_items; i>1; i-) bool exchange=false; for (int j=1; j<

27、;i; j+) if (pexpsj < pexpsj-1) /交換pexpsj和pexpsj-1 int temp1=pexpsj; pexpsj = pexpsj-1; pexpsj-1 = temp1; /交換pcoefsj和pcoefsj-1 double temp2 = pcoefsj; pcoefsj = pcoefsj-1; pcoefsj-1 = temp2; exchange = true; if (!exchange) break; Polynomial:Polynomial(const Polynomial &p)num_of_items=p.num_of_

28、items;pcoefs=new doublenum_of_items;pexps=new intnum_of_items;for (int i=0; i<num_of_items; i+)pcoefsi=p.pcoefsi;pexpsi=p.pexpsi;Polynomial:Polynomial()delete pcoefs;delete pexps; pcoefs=NULL; pexps=NULL; num_of_items=0;Polynomial& Polynomial:operator=(const Polynomial &p)delete pcoefs; d

29、elete pexps; num_of_items=p.num_of_items;pcoefs=new doublenum_of_items;pexps=new intnum_of_items;for (int i=0; i<num_of_items; i+)pcoefsi=p.pcoefsi;pexpsi=p.pexpsi;return *this;int Polynomial:degree() constif (num_of_items = 0) return 0; else return pexpsnum_of_items-1;double Polynomial:evaluate(

30、double x) constdouble sum=0;for (int i=0; i<num_of_items; i+)double temp=pcoefsi;for (int j=0; j<pexpsi; j+)temp *= x;sum += temp;return sum;bool Polynomial:operator=(const Polynomial &p) constif (num_of_items != p.num_of_items) return false;for (int i=0; i<num_of_items; i+)if (pcoefsi!

31、=p.pcoefsi | pexpsi!=p.pexpsi)return false;return true;bool Polynomial:operator!=(const Polynomial &p) const return !(*this = p);Polynomial Polynomial:operator+(const Polynomial &p) constdouble *coefs=new doublenum_of_items+p.num_of_items;int *exps=new intnum_of_items+p.num_of_items;int coun

32、t=add(p,coefs,exps);Polynomial temp(coefs, exps, count); delete coefs; delete exps;return temp;Polynomial Polynomial:operator-(const Polynomial &p) constdouble *coefs=new doublenum_of_items+p.num_of_items;int *exps=new intnum_of_items+p.num_of_items;int count=subtract(p,coefs,exps);Polynomial te

33、mp(coefs, exps, count); delete coefs; delete exps;return temp;Polynomial Polynomial:operator*(const Polynomial &p) const Polynomial sum,temp=*this; for (int i=0; i<p.num_of_items; i+) for (int j=0; j<num_of_items; j+) temp.pcoefsj = pcoefsj * p.pcoefsi; temp.pexpsj = pexpsj + p.pexpsi; sum

34、 += temp; return sum;Polynomial &Polynomial:operator+=(const Polynomial &p)double *coefs=new doublenum_of_items+p.num_of_items;int *exps=new intnum_of_items+p.num_of_items;int count=add(p,coefs,exps); delete pcoefs; delete pexps; pcoefs = coefs; /有多余的空間，實際的項數(shù)由count決定。 pexps = exps; /同上。 num_

35、of_items = count;return *this;Polynomial &Polynomial:operator-=(const Polynomial &p)double *coefs=new doublenum_of_items+p.num_of_items;int *exps=new intnum_of_items+p.num_of_items;int count=subtract(p,coefs,exps); delete pcoefs; delete pexps; pcoefs = coefs; /有多余的空間，實際的項數(shù)由count決定。 pexps = e

36、xps; /同上。 num_of_items = count;return *this;Polynomial &Polynomial:operator*=(const Polynomial &p)Polynomial sum,temp=*this; for (int i=0; i<p.num_of_items; i+) for (int j=0; j<num_of_items; j+) temp.pcoefsj = pcoefsj * p.pcoefsi; temp.pexpsj = pexpsj + p.pexpsi; sum += temp; *this = s

37、um;return *this;6、 用操作符重載重新實現(xiàn)習題6-8中的第17題集合類的一些操作：<=（包含于）、=（相等）、!=（不等）、 |（并集）、 &（交集）、-（差集）、+=（增加元素）、-=（刪除元素）等。解：#include <iostream>using namespace std;struct Nodeint value;Node * next;class IntSetpublic:IntSet();IntSet(const IntSet& s);IntSet();bool is_empty() const; /判斷是否為空集。int siz

38、e() const; /獲取元素個數(shù)。bool is_element(int e) const; /判斷e是否屬于集合。void display() const; /顯示集合中的所有元素。bool operator <=(const IntSet& s) const; /判斷s是否包含于集合。bool operator =(const IntSet& s) const; /判斷集合是否相等。bool operator !=(const IntSet& s) const; /判斷集合是否不相等。IntSet& operator +=(int e); /將e加

39、入到集合中。IntSet& operator -=(int e); /把e從集合中刪除。IntSet operator |(const IntSet& s) const; /計算集合的并集。IntSet operator &(const IntSet& s) const; /計算集合的交集。IntSet operator -(const IntSet& s) const; /計算集合的差。 IntSet &operator = (const IntSet& s); /集合賦值private:int count;Node *head;Int

40、Set:IntSet()count=0;head=NULL;IntSet:IntSet(const IntSet &s)head = NULL;count = 0;Node *p=s.head;for (int i=0; i<s.count; i+)*this += (p->value);p = p->next;IntSet:IntSet()Node *p;while (head!=NULL)p = head;head = head->next;delete p;count=0;bool IntSet:is_empty() constreturn count=0

41、?true:false;int IntSet:size() constreturn count;bool IntSet:is_element(int e) constfor (Node *p=head;p!=NULL;p=p->next)if (p->value = e) return true;return false;bool IntSet:operator <=(const IntSet& s) constfor (Node *p=s.head;p!=NULL;p=p->next)if(!is_element(p->value) return fal

42、se;return true;bool IntSet:operator =(const IntSet& s) constif (count!=s.count)return false;else if (*this <= s && s <= *this)return true;elsereturn false;bool IntSet:operator !=(const IntSet& s) const return !(*this = s);void IntSet:display() constfor (Node *p=head;p!=NULL;p=p

43、->next)cout << p->value << 't'cout << endl;IntSet& IntSet:operator +=(int e)if(!is_element(e)Node *p=new Node;p->value = e;p->next = head;head = p;count+;return *this;IntSet& IntSet:operator -=(int e)if(is_element(e)if (head->value=e)Node *p=head;head

44、 = head->next;delete p;elsefor (Node *p=head; p->next!=NULL; p=p->next)if (p->next->value=e)Node *temp=p->next;p->next = temp->next;delete temp;break;count-;return *this;IntSet IntSet:operator |(const IntSet& s) constIntSet set(s);Node *p=head;while (p!=NULL)if (!set.is_e

45、lement(p->value)set += p->value;p=p->next;return set;IntSet IntSet:operator &(const IntSet& s) constIntSet set;Node *p=head;while (p!=NULL)if (s.is_element(p->value)set += p->value;p = p->next;return set;IntSet IntSet:operator -(const IntSet& s) constIntSet set;Node *p=

46、head;while (p!=NULL)if (!s.is_element(p->value)set += p->value;p = p->next;return set;IntSet &IntSet:operator = (const IntSet& s) Node *p;while (head!=NULL)p = head;head = head->next;delete p;count=0; p=s.head;for (int i=0; i<s.count; i+)*this += (p->value);p = p->next;

47、return *this;7、 定義一個帶下標范圍檢查、數(shù)組整體賦值和比較功能的一維int型數(shù)組類：IntArray。解：#include <iostream>using namespace std;class IntArrayint *p_buf;int size;public:IntArray(int n)size = n;p_buf = new intsize;for (int i=0; i<size; i+)p_bufi = 0;IntArray(const IntArray& x)size = x.size;p_buf = new intsize;for (

48、int i=0; i<size; i+)p_bufi = x.p_bufi;IntArray()delete p_buf;p_buf = NULL;size = 0;int& operator(int i) /取某位置上的字符，if (i < 0 | i >= size) cerr << "超出數(shù)組范圍!n"exit(-1);return p_bufi;int operator(int i) const /取某位置上的字符，用于常量對象if (i < 0 | i >= size) cerr << "超出

49、數(shù)組范圍!n"exit(-1);return p_bufi;IntArray& operator =(IntArray& x) if (&x = this) return *this;if (size != x.size) cerr << "兩個數(shù)組尺寸不同!n"exit(-1);for (int i=0; i<size; i+)p_bufi = x.p_bufi;return *this;bool operator =(IntArray& x)if (&x = this) return true;for

50、(int i=0; i<size; i+)if (p_bufi != x.p_bufi) return false;return true;bool operator !=(IntArray& x)return !(*this = x); 8、 定義一個不受計算機字長限制的整數(shù)類INT，要求INT與INT以及INT與C+基本數(shù)據(jù)類型int之間能進行、×、÷和運算，并且能通過cout輸出INT類型的值。解：#include <iostream>#include <cstring>using namespace std;class INTc

51、har *p_buf; /從低位到高位存儲整型數(shù)，int buf_len, /p_buf所占空間大?。?0的倍數(shù)） sign; /整型數(shù)的符號（1為正，-1為負）void Add(const INT &i); /把i加到*this中void Minus(const INT &i); /從*this中減去iINT(char *p, int len, int s=1); /新對象的p_buf在已有的空間p上（不再另外分配空間）。public :INT();INT(int i);INT(char *num);INT(const INT &i);INT();INT &operator=(const INT &i); INT& INT:operator+=(const INT &i); INT& INT:operator-=(const INT &i);friend INT operator+(const INT &i1, const INT &i2);friend INT operator-(const INT