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Structure and User-Define Type

Structure and User-Define Type. struct tag { members;…;} variable; typedef old_type new_type ;. Structures 結構變數. Syntax: struct type_id { type1 variable1; type2 variable2; …} varible1, ..; struct type_id variable1, variable2;. 4 bytes. 8 bytes. &a. &b. &ss.

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Structure and User-Define Type

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  1. Structure and User-Define Type structtag {members;…;} variable; typedef old_type new_type;

  2. Structures 結構變數 • Syntax: structtype_id { type1 variable1; type2 variable2; …} varible1, ..; structtype_id variable1, variable2;

  3. 4 bytes 8 bytes &a &b &ss 4 bytes 8 bytes &ss.n &ss.x Memory layout 計憶體的配置 • int a; double b; • struct {int n; double x;} ss;

  4. Structure members (example)結構變數之成員 struct item_record { unsigned int number; float price; } pen, book; pen.number = 10; pen.price = 13.5; book.number = 5; book.price = 590.5; total = pen.price * pen.number + book.price * book.number; struct item_record notebook; notebook.price = 60,4; notebook.number = 20; total = total + notebook.price * notebook.number;

  5. User-Type Define (example) 自定新的變數型態 Syntax: (語法) typedef old_type_name new_type_name; typedef unsigned int Uint; typedef struct struct_type new_type_name;

  6. Example 2. 將相關變數集合為一變數結構 typedef struct { char name[80]; unsigned long int id; float t1, t2, t3, avg;} Student; Student sss; // Student 為一新的變數型態 sss.name // name of student. 姓名 sss.id // id number. 學號 sss.t1 // score of first test. 成績 …. sss.avg // average over t1, t2, and t3 平均

  7. While Loop 用 while 控制迴圈 (1) while (condition) { …; …..; …..; } (2) do { …; …; …; …; } while (condition); sum = 0; n = 0; while (n <= 10) { sum = sum + n; n++; } char a; printf(“Press q to quit”;) fflush(stdin); do { a = getc(stdin); } while (a != ‘q’);

  8. File input: getc(file_pointer)從檔案中讀取一個字元 Symbol represents the file pointer of Keyboard: stdin : standard input device, defined within stdio.h getc(stdin) : read a single character from keyboard.

  9. 直角座標 與 球座標 轉換 ex4 typedef struct { double x, y, z;} Cartesian; typedef struct { double r, t, p;} Spherical; Cartesian v1; Spherical v2; v2.r = sqrt(v1.x*v1.x + v1.y*v1.y+v1.z*v1.z); v2.t = acos(v1.z / v2.r); v2.p = atan(v1.y / v1.x);

  10. Practice 2. 增加 expval(x) 的可用範圍 A double variable can represent a value < ~ exp(709) exp(800) = #INF 超出 double 的範圍 Problem : How to increase the applicable range of our previous program?

  11. Numerical Scheme 選擇 a = ln(10) = 2.302585092994045684 選擇 n 使 y=(x – na) , 0 <= y < ln(10)

  12. Flow Chart of the Scheme mytype myexp() main double expval() Input x myexp(x) expval(y) nq = (int)floor(x / LN10) Set n = 30 Call myexp(x) Do Taylor series In reverse order y = x – nq *LN10 Output result return sum return expval(y), nq

  13. Using Netwon’s method find the root of Pratice 5. Another root searching Write the function and its derivative within a same subroutine, And return these values in a two-double structure.

  14. details struct two_double { double x; double y;}; typedef struct two_double twodim; twodim funct(double x) { twodim fdf; ……. fdf.x = ??; fdf.y = ??; return fdf; } int main() { twodim ff; …. do { … ff = funct(x); delta = ff.x / ff.y; } …… }

  15. Related usages or functions • #include <math.h> double floor(double a) ; // return the largest integer that smaller than a. • typedef struct {double x; int n}MyType; For example : floor(2.34) = 2.00, floor(-3.48) = -4.00 MyType myexp(x) { MyType ttt; ……; …..; ttt.n = nq; ttt.x = expval(y); return ttt; }

  16. Project 3. 計算二次方程的根 設計一程式 • 寫一函數讀取系數 a, b, c. 回傳主程式. • 將系數傳到令一函數計算兩複數根. • 將兩複數根傳回主程式 • 再傳到另一副程式列印結果.

  17. 自定新變數格式 • Quardratic typedef struct {double a, b, c;} Quardratic; • Complex typedef struct {double x; double y;} Complex; • TwoComplex typedef struct { Complex r1; Complex r2; } TwoComplex;

  18. functions • 讀取系數, 並回傳到主程式 Quadratic inputcoef(void); • 給與系數計算兩根 TwoComplex findroots(Quadratic coef); • 列印 void printroots(TwoComplex root);

  19. Techniques -- input • Quadratic inputcoef(void) { double a, b, c; Quadratic coef; …. 讀入 a, b, c… coef.a = a; coef.b = b; coef.c = c; return coef; }

  20. Output viod printroots (TwoComplex twor) { printf(“root1 = (%.10lf, %.10lf)\n”, twor.r1.x, twor.r2.y); ……… return; }

  21. Main and prepocessors #include <stdio.h> #include <stdlib.h> #include <math.h> typedef struct {double a, b, c;} Quadratic; typedef struct {double x, y;} Complex; typedef struct {Complex r1, r2;} TwoComplex; Quadratic inputcoef(void); TwoComplex findroots(Quadratic); void printroots(TwoComplex); int main() { Quadratic coefs; TwoComplex roots; coefs = inputcoef(); roots = findroots(coefs); printroots(roots); system("pause"); return 0; } 輸入系數 解根 列印

  22. Complex root1, root2; TwoComplex root; a  cf.a b  cf.b c  cf.c findroot(Quadratic cf) 判別式 crit = b*b – 4*a*c crit >= 0.0 ? no yes crit = sqrt(-crit) crit = sqrt(crit) root1.x = (-b+crit) / (2a) root2.x = (-b-crit) / (2a) root1.y = root2.y = 0.0; root1.x = root2.x = (-b) / (2a) root1.y = crit / (2a) root2.y = (-crit) / (2a); root.r1 = root1; root.r2 = root2; return root;

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