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1、P. 1C+的多面性(Overloading)o 使一個算符或函式具有處理多種資料型別能力的方法稱為複載(Overloading)o C+中對複載的意義為:讓同一種名稱或用法具有多種意義o 當我們定義多個具有相同名稱,但卻m 有不同參數個數或m 相同參數個數但參數型別不同 時, 這就稱為函式的複載 函式及運算子函式及運算子的多面性的多面性 (Overloading) 函式函式的複載的複載 (Overloading)int max(int a, int b) .char max(char a, char b) .long max(long a,long b) .int i=max(2,4);ch
2、ar c=max(a,b);long l=max(23L,56L);P. 2C+的多面性(Overloading)o 函式傳回值的型別以及 函式的參數名稱不可作 為複載函式的識別之用int print();long print(); / errorint print(int a);int print(int b); / errorTypedef char flag;print(char);print(flag); / erroro 不同的 scope 各有其獨自的 overloading 空間Func1() int max(int a, int b); . Func2() int max(ch
3、ar a, char b); . o由於 typedef 並不會定義出新的型別 (只是製造出一別名而已),因此,用 typedef 所定義出的型別仍以其原始型別為複載之 依據P. 3C+的多面性(Overloading)/ Overloading 使用範例#include void repchar();void repchar(char0;void repchar(char, int);void main() repchar(); repchar(=); repchar(+,30);void repchar() for ( int j=0;j45;j+) cout *; cout endl;v
4、oid repchar(char ch) for (int j=0;j45;j+) cout ch; cout endl;void repchar(char ch, int n) for (int j=0; jn; j+) cout ch; cout endl;Output:*=+ +P. 4C+的多面性(Overloading)/ Overloading 使用範例#include #include #include /for exit(1)const int Max=100;class Stack protected: int stMax; int top; public: Stack()t
5、op=0; void Push(int var) st+top=var; int Pop() return sttop-; ;class Stack2: public Stack public: void Push(int var) if (topMax) Stack:Push(var); else cout 0) return Stack:Pop(); else cout nError: stack is empty; exit(1); ;void main() clrscr(); Stack2 s2; s2.Push(11); s2.Push(22); s2.Push(33); cout
6、endl s2.Pop(); cout endl s2.Pop(); cout endl s2.Pop(); cout endl ,=.)以及 算術指定運算子(+=,*=,) 能對基本資料型態如 int, float, long 等執行運算 如: int a=b+c; leage+=le;o對使用者自行定義較複雜的資料型態如結構或類別,就不能拿這些運算子 直接作運算,例如,不能直接將兩個屬於 room 的類別變數作相加來當作其 成員變數相加,必須另外透過成員函數進行運算. .Class room private: float ledge, sedge; public: void addsqua
7、re(room r1, room r2) r1, r2, r3;.void room : addsquare(room r1, room r2) ledge=r1.ledge + r2.ledge; sedge=r1.sedge+ r2.sedge;void main() . r3.addsuare(r1, r2); r3=r1+r2; /error P. 6C+的多面性(Overloading)o C+提供運算子的複載(多元定義 operator overloading), 允許使用者對 運算子重新定義, 經重新定義的運算子也能對複雜的資料型態進行類 似的運算如 r3=r1+r2;o 對使用
8、者自行定義較複雜的資料型態如結構或類別,就不能拿這些運算子 直接作運算,例如,不能直接將兩個屬於 room 的類別變數作相加來當作其 成員變數相加,必須另外透過成員函數進行運算. #include / page 12-5/ 單一運算元:無參數,無 return值class incount private: int c1,c2; public: incount() c1=0; c2=1000; void retcount(void) cout c1= c1 endl; cout c2= c2 endl;void operator +() c1+; c2+; ;void main() incoun
9、t ic1,ic2; ic1.retcount(); ic2.retcount(); ic1+; / considered as ic1.+() ic2+; / considered as ic2.+() ic1.retcount(); ic2.retcount(); c1c2ic1c1c2ic1c1c2ic2P. 7C+的多面性(Overloading)#include / page 12-8/ 單一運算元:無參數,有 return值 class incount private: int c1,c2; public: incount() c1=0; c2=1000; void retcoun
10、t(void) cout c1= c1 endl; cout c2= c2 endl; incount operator +() c1+; c2+; incount temp; temp.c1=c1; temp.c2=c2; return temp; ;void main() incount ic1,ic2; ic1.retcount(); ic2.retcount(); ic1+; / considered as ic1.+ ic1.retcount(); ic2=ic1+; ic2+.retcount(); / considered as ic2.+ Ans:c1=0 c2=1000 c1
11、=0 c2=1000 c1=1 c2=1001 c1=3 c2=1003#include / page 12-11/ 單一運算元:無參數,有 return值 class incount private: int c1,c2; public: incount() c1=0; c2=1000; incount(int vc1, int vc2) /overloading c1=vc1; c2=vc2; void retcount(void) cout c1= c1 endl; cout c2= c2 endl; incount operator +() c1+; c2+; / unnamed ob
12、ject initialized return return incount(c1,c2); ;void main() incount ic1,ic2; ic1.retcount(); ic2.retcount(); ic1+; ic1.retcount(); ic2=ic1+; ic2+.retcount(); Ans:c1=0 c2=1000 c1=0 c2=1000 c1=1 c2=1001 c1=3 c2=1003P. 8C+的多面性(Overloading)o 物件相加多元運算 +o 欲以運算子定義及多元運算設計使物件可直接相加#include class room /Page:12
13、-14 private: float ledge,sedge; public: room() ledge=0.0; sedge=0.0; room(float le, float se) ledge=le; sedge=se; void getlength() cout ledge;cout sedge; void showsquare() cout (ledge+sedge)*2 endl; room operator + (room p2); room room:operator + (room p2) float led=ledge+p2.ledge; float sed=sedge+p
14、2.sedge; return room(led, sed); void main() room r2; room r1(3,2); r2.getlength(); cout Length of r1 room is:; r1.showsquare(); cout Length of r2 room is:; r2.showsquare(); room r3=r1+r2; cout Length of r3 room is:; r3.showsquare(); room r4=r1+r2+r3; cout Length of r4 room is:; r4.showsquare();P. 9C
15、+的多面性(Overloading)o 物件相加多元運算 +#include /Page:12-18#include #include const PI=3.14159;class rectanglar private: double x,y; double getr() return sqrt(x*x+y*y); double getangle() return atan(y/x)*180/PI; public: rectanglar() x=0;y=0; rectanglar(double p, double q) x=p; y=q; void display1() cout ( x ,
16、y )=Polar; cout ( setw(5) setprecision(2) getr() , getangle() ); rectanglar operator + (rectanglar r2) double p=x+r2.x;double q=y+r2.y;return rectanglar(p,q); ;void main() rectanglar r1(20,10); rectanglar r2(15,20); rectanglar r3=r1+r2; /considered as r1.+(r2) rectanglar r4=r1+r2+r3; / as r1.+(r2.+(
17、r3) cout n rectanglar r1;r1.display1(); cout n rectanglar r2;r2.display1(); cout n rectanglar r3;r3.display1(); cout n rectanglar r4;r4.display1();xy2010r1xy1520r2xyr3xyr4xyRectanglar (p,q)P. 10C+的多面性(Overloading)#include class Distance private: int feet; float inches; public: Distance() feet=0; inc
18、hes=0; Distance(int ft, float in) feet=ft; inches=in; void getdist() cout feet; cout inches; void showdist() cout feet - inches =12.0) i-=12.0; f+; return Distance(f,i);void main() Distance dist1,dist3,dist4; dist1.getdist(); Distance dist2(11,6.25); dist3=dist1+dist2; dist4=dist1+dist2+dist3; cout
19、ndist1=;dist1.showdist(); cout ndist2=;dist2.showdist(); cout ndist3=;dist3.showdist(); cout ndist4=;dist4.showdist();P. 11C+的多面性(Overloading)o 物件相加多元運算 :=#include class room /Page:12-20 private: float ledge,sedge; public: room() ledge=0.0; sedge=0.0; room(float le,float se) ledge=le; sedge=se; void
20、 getlength() cout ledge; cout sedge; void showsquare() cout (ledge+sedge)*2 endl; room operator += (room p2);room room:operator += (room p2) ledge +=p2.ledge; sedge +=p2.sedge; return room(ledge, sedge);void main() room r2; room r1(8,7); r2.getlength(); cout Length of r1 is:;r1.showsquare(); cout Le
21、ngth of r2 is:;r2.showsquare(); room r3=r1+=r2; cout Length of r3 is:;r3.showsquare(); room r4=r1+=r3; cout Length of r4 is:;r4.showsquare();sedgeledge8.07.0r1sedgeledge12.08.0r2sedgeledger3sedgeledgeroom(ledge,sedge)P. 12C+的多面性(Overloading)o 物件相加多元運算 : 字串相加#include #include const int size=80;class
22、strings /Page:12-24 private: char strsize; public: strings() str0=0; strings(char st) strcpy(str,st); void printstr() cout str; strings operator + (strings p);strings strings:operator + (strings p) if (strlen(str)+strlen(p.str)size) strings ptemp; strcpy(ptemp.str,str); strcpy(ptemp.str,p.str); return ptemp; else cout n string limited 80 chars; void main() strings ps1=n London bridge ; strings ps2=is falling down !; ps1.printstr(); cout endl; ps2.printstr(); strings ps3=ps1+ps2; ps3.printstr();
