Microsoft Word - Prog1-981.docx

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畴安
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1 5. 變數參照 (Memory Reference) 5.1 指標 (Pointer) (1). 指標 (Pointer) 的基本觀念特性內含為一 Memory Address 會因不同的機器而有不同的結果 &" 也是代表變數的位址例如 : int var1 = 2; cout << var1 << endl; cout << &var1 << endl; 會印出 : 2,0x22ff74 其中 2 為 var1 的值 (value),0x22ff74 為 var1 所存放的位置 (address) address & value int var1; int *var2; var2 為儲存 address 的變數 *var2 則指該位址真正儲存的值所以 var2 = &var1 例如 : int var1 = 12; int *var2 = &var1; *var2 = 13; cout << var1 << "," << var2 << endl; 會印出 : 13,0x22ff74 其中 13 為 var1 的值 (value), 而 var1 的位址則存放於 var2 中, 0x22ff74 為 var1 所存放的位置 (address) 練習 : var2++ 代表指向下一個位址 string 也可使用 pointer 的方式來宣告 char name[] = " "; char *var = name; cout << "(1)=>" << var << endl; var++; cout << "(2)=>" << var << endl; 68

2 var++; cout << "(3)=>" << var << endl; cout << "*var=>" << *var << endl; 印出 : (1)=> (2)=> (3)=> *var=>3 上面的例子也可以用下式來指定 var1 的值 char *var = ; *var1 則只是指該位置上的 char 而已下面程式執行後的結果為何? int var1 = 18; int* var2 = &var1; *var2 = 12; cout << var1 << endl; Ans: 12 下面程式執行後的結果為何? 1 int var1 = 18; 2 int* var2 = &var1; 3 var2 = 12; 4 cout << var1 << endl; Ans: 第 3 行錯誤, 正確為 *var2 = 12; 下面程式執行後的結果為何? int var1 = 11; test1(&var1); cout << var1 << endl; void test1(int *v1){ if (*v1>10) *v1 = *v1+1; *v1 = *v1+1; Ans: 13 char *var = " "; cout << var << endl; 69

3 Ans: char *var = " "; cout << *var << endl; Ans: 1 char *var = " "; cout << *var; var++; cout << *var << endl; Ans: 12 int var1 = 12, var2 = 22; int *var21 = &var1; int *var22 = &var2; int **var31 = &var21; int **var32 = &var22; **var32 = **var31+**var32; cout << **var32 << endl; Ans: 34 (2). 函數參數的傳遞 Call by value 為一般的 Function Call, Function 並不會改變參數的值當 Function 接到參數後, 會在 Memory 中複製一份 Example: void main(){ int a = 12; test1(a); cout << a << endl; // a = 12; void test1(int a){ a = 0; 70

4 結果印出 : 12 Call by Reference 主程式所傳給 Function 的參數為變數的位址當 Function 接到參數後, 並不會在 Memory 中複製一份所以當 Function 改變變數內容後, 主程式的變數也一併改變了 Example: void main(){ int a = 12; test(&a); cout << a << endl; // a = 0; void test1(int *a){ *a = 0; 結果印出 : 0 int var1 = 11; test1(var1); cout << var1 << endl; void test1(int v1){ v1 = 12; Ans: 11 int var1 = 11; test1(&var1); cout << var1 << endl; void test1(int *v1){ *v1 = 12; Ans: 12 int var1 = 11, var2 = 12; test1(&var1, var2); cout << var1 << "," << var2 << endl; 71

5 void test1(int *v1){ *v1 = *v1+v2; v2 = *v1+1; Ans: 23,12 int var1 = 11, var2 = 12; test1(&var1, var2); test2(var1, &var2); cout << var1 << "," << var2 << endl; void test1(int *v1, int v2){ *v1 = *v1+v2; v2 = *v1+v2; void test2(int v1, int *v2){ v1 = v1+*v2; *v2 = v1+*v2; Ans: 23,47 int var1 = 11, var2 = 12; test1(&var1, test2(var1,&var2)); cout << var1 << "," << var2 << endl; void test1(int *v1, int v2){ *v1 = *v1+v2; v2 = *v1+v2; int test2(int v1, int *v2){ v1 = v1+*v2; *v2 = v1+*v2; return v1; Ans: 34,35 72

6 5.2 結構 (Structure) (1). 基本結構性態特性主要的目的是為了要表示一些在 C++ 內沒有定義的資料型態所表式的資料結構可以是 Recursive 使用方式 struct Struc_Name { Data_Type var_name; Data_Type var_name;... ; struct Struc_Name my_struc; Struc_Name: 表示結構的名稱 Data_Type: 為變數的名稱在結束的的地方要加上 ;" 要使用自己所定義的 Structure, 需要採用和一般變數宣告的方式一樣 struct Struc_Name my_struc; Example: struct Point { int x; int y; ;... struct Point pt1; pt1.x = 120; pt1.y = 200; cout << pt1.x <<, << pt1.y << endl; 則印出 : 120,200 Structure 的初值設定也可以利用 struct Point pt1 = {120,200; 語法 : 直接使用變數宣告, 則用 "." 來存取 struct Person { char name[10]; ; struct Person per1;... per1.id = 5; 73

7 strcpy(per1.name,"mary"); 若宣告為 Pointer 則用 "->" 存取 struct Person* per2; per2->id = 5; strcpy(per2->name, "Mary"); Structure & Function 在函數間傳 Structure 和傳一般參數相同 Point pt1={100,200, pt2; pt2 = add3(pt1);... Point add3(point pt) {... 要注意的是函數的 Prototype 要在所定義的 Structure 之後 (2). 範例練習 struct Person { char name[10]; ; struct Person per1 = {1, test1, per2 = {2, test2 ; per1.id = 3; cout << per1.id <<, << per2.id << endl; Ans: 3,2 struct Person { char name[10]; ; struct Person per1[3]; void setperson(int, int, char*); setperson(0, 3, Mary ); setperson(1, 5, John ); setperson(2, 8, Peter ); cout << per1[1].id <<, << per2[2].name << endl; void setperson(int v0, int v1, char* v2){ 74

8 per1[v0].id = v1; strcpy(per1[v0].name,v2); Ans: 5,Peter struct Product { char name[10]; int price; ; struct Product pro[2]; void setproduct(int, int, char*, int); setproduct (0, 3, Book,50); setproduct (1, 5, Table,800); cout << pro[1].id <<, << pro[1].price << endl; void setproduct (int v0, int v1, char* v2, int v3){ pro[v0].id = v1; strcpy(pro[v0].name,v2); pro[v0].price = v3; Ans: 1,800 struct Person { char name[10]; Person* next; ; struct Person per; void setperson(person*, Person *, int, char*); void setperson(person* p0, int v1, char* v2){ p0->id = v1; strcpy_s(p0->name,v2); void main(){ setperson(&per, 11, "Mary"); struct Person per1; setperson(&per1, 21, "Peter"); per.next = &per1; struct Person* per2 = &per; per2 = per2->next; 75

9 cout << per2->id << "," << per2->name << endl; system("pause"); Ans: 21,Peter 範例 struct Person { Person* next; ; struct Person per; void main(){ struct Person per2, per3; per1.id = 11; per1.next = &per2; per2.id = 21; per2.next = &per3; per3.id = 31; struct Person* per = &per1; per = per->next; cout << per->id << ","; per = per->next; cout << per->id << endl; Ans: 21, 綜合練習 (1). 貨品管理程式 (product_struct) 本練習要撰寫一個進出貨管理程式程式需求如下 : 需要利用結構方式 (structure) 來實作每項貨品 (product) 需要有以下屬性 (1) 貨號 (id) (2) 品名 (name) (3) 進價 (pricein) (4) 售價 (price) (5) 庫存數量 (amount) 需要提供一個選單式的介面, 並具有 : (1) (L) 列出所有貨品 (2) (N) 新增品貨品 76

10 (3) (D) 刪除貨品 ( 最後一筆 ) (4) (F) 由貨號查詢貨品 (5) (Q) 離開程式按下指定的鍵來執行相關功能每項功能顯示前需要先清除螢幕可利用 Link list 的方式來增加貨品管理的彈性 77

Microsoft Word - Java-982.docx

Microsoft Word - Java-982.docx 4. 類別 (class) 4.1 類別 (Class) 與物件 (Object) (1). 基本觀念相同特性的變數在命名時會造成不方便例如利用 name 來表示名稱, 則 : 桌子的名稱 => String TableName = " 辦公桌 "; 椅子的名稱 => String ChairName = " 辦公椅 "; 因此不同物件的名稱皆需要加入前置字來區別 ( 十分不方便 ) C 語言使用