1、,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)C+中對複載的意義為:讓同一種名稱或用法具有多種意義當我們定義多個具有相同名稱,但卻有不同參數個數或相同參數個數但參數型別不同時, 這就稱為函式的複載,函式及運算子的多面性 (Overloading),函式的複載 (Overloading),int max(int a, int b) . char max(char a, char b) . long max(long a,long b) .int i=max(2,4); char c=max(a,b); long l=max(23L,56L);,函式傳回值的型別
2、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); / error,Typedef char flag;print(char); print(flag); / error,不同的 scope 各有其獨自的 overloading 空間,Func1() int max(int a, int b);. ,Func2() int max(char a, char b);. ,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
3、定義出的型別仍以其原始型別為複載之依據,/ 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; ,void repchar(char ch) for (int j=0;j45;j+)cout ch;cout endl; void repchar(char ch, int n)
4、for (int j=0; jn; j+)cout ch;cout endl; Output: * = + +,/ Overloading 使用範例#include #include #include /for exit(1) const int Max=100; class Stack protected:int stMax;int top;public:Stack()top=0;void Push(int var) st+top=var; int Pop() return sttop-; ;,class Stack2: public Stack public:void Push(int v
5、ar) if (top0)return Stack:Pop();else cout “nError: stack isempty“;exit(1); ;,void main() clrscr();Stack2 s2;s2.Push(11);s2.Push(22);s2.Push(33);cout endl s2.Pop();cout endl s2.Pop();cout endl s2.Pop();cout endl s2.Pop(); ,Output: 33 22 11 Error: stack is empty,運算子的多面性 (Operator Overloading),C+與C相同,提
6、供算術運算子(+,-,*,/,+,-)及關係運算子(,=)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;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 addsquare(room r1, room r2) r1, r2, r3; .,void room : a
7、ddsquare(room r1, room r2) ledge=r1.ledge + r2.ledge;sedge=r1.sedge+ r2.sedge; void main() . r3.addsuare(r1, r2);r3=r1+r2; /error ,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;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
8、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() incount ic1,ic2;ic1.retcount();ic2.retcount();ic1+; / considered as ic1.+()ic2+;
9、 / considered as ic2.+()ic1.retcount();ic2.retcount();,#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 retcount(void) cout “c1=“ c1 endl;cout “c2=“ c2 endl;incount operator +() c1+; c2+;incount temp;temp.c1=c1; temp.c2=c2;return temp; ;
10、 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=1000c1=0 c2=1000c1=1 c2=1001c1=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;
11、 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 object initialized returnreturn incount(c1,c2); ; void main() incount ic1,ic2;ic1.retcount(); ic2.retcount();ic1+; ic1.retcount();ic2=ic1
12、+;ic2+.retcount();,Ans:c1=0 c2=1000c1=0 c2=1000c1=1 c2=1001c1=3 c2=1003,物件相加多元運算 +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:12-14 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(
13、) cout (ledge+sedge)*2 endl;room operator + (room p2); ;,room room:operator + (room p2) float led=ledge+p2.ledge;float sed=sedge+p2.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();r
14、oom r3=r1+r2;cout “Length of r3 room is:“;r3.showsquare();room r4=r1+r2+r3;cout “Length of r4 room is:“;r4.showsquare(); ,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
15、;public:rectanglar()x=0;y=0;rectanglar(double p, double q)x=p; y=q;void display1() cout “(“ x “,“ 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);r
16、ectanglar r2(15,20);rectanglar r3=r1+r2; /considered as r1.+(r2)rectanglar r4=r1+r2+r3; / as r1.+(r2.+(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(); ,x,y,Rectanglar (p,q),#include class Distance
17、 private:int feet;float inches;public:Distance() feet=0; inches=0; Distance(int ft, float in) feet=ft; inches=in; void getdist() cout feet;cout inches; void showdist() cout feet “-“ inches “; Distance operator + (Distance); ;,Distance Distance: operator + (Distance d2) int f=feet+d2.feet;float i=inc
18、hes+d2.inches;if (i=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 “ndist1=“;dist1.showdist();cout “ndist2=“;dist2.showdist();cout “ndist3=“;dist3.showdist();cout “ndist4=“;dist4.sh
19、owdist(); ,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 getlength() cout ledge;cout sedge;void showsquare() cout (ledge+sedge)*2 endl;room operator += (room p2); ; room room:operator += (room p2
20、) 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 “Length 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()
21、; ,sedge,ledge,room(ledge,sedge),物件相加多元運算 : 字串相加,#include #include const int size=80; class 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(); ,
