1. 13 주 강의 Structures and Unions

2. Structures Structure – 기본 자료형들로 구성된 새로운 자료형 예 ) struct card { intpips; char suit; };

3. Structures  struct card { int pips; char suit; }; struct cardc1, c2;  struct card { int pips; char suit; } c1, c2; ※ pips,suit 를 멤버로 가지는 card Type 의 c1,c2 선언

4. 값 저장 c1.pips = 3; c1.suit = ‘ s ’ ; c2=c1; /* structure 전체를 복사 */ typedef struct card card; card c3, c4, c5;

5. 다른 예 struct fruit { char *name; int calories; } struct vegetable { char *name; int calories; } struct fruit a; struct vegetable b;

6. 예 struct card { int pips; char suit; } deck[52]; struct date { int day, month, year; char day_name[4]; /* Mon, Tue,.. */ char month_name[4]; /* Jan, … */ } yesterday, today, tomorrow;

7. Complex number typedef struct { float re; float im; } complex; complex a, b, c[100];

8. Accessing members #define CLASS_SIZE 100 struct student { char *last name; int student_id; char grade; }; struct student tmp, class[CLASS_SIZE]; tmp.grade = ‘ A ’ ; tmp.last_name = “ Casanova ” ; tmp.student_id = 910017; class[i].grade = ‘ F ’ ;

9. 예제 설명 및 -> Page 412 프로그램 설명 -> – pointer_to_structure->member_name  (*pointer_to_structure).member_name /* () 가 없으면 오류 */ complex *a, *b, *c; – a->re = b-> re + c -> re;

10. Structures structure_declaration::=struct_specifier declarator_list; struct_specifier::=struct tag_name struct tag_name opt {{member_declaration} 1+ } tag_name::=identifier member_declaration::=type_specifier declarator_list; declarator_list::=declarator{,declarator} 0+

11. Member access operators Declarations and assignments struct student tmp, *p = &tmp; tmp.grade = ‘ A ’ ; tmp.last_name = “ Casanava ” ; tmp.student_id = 910017; ExpressionEquivalent expressionConceptual value tmp.gradep->gradeA tmp.last_namep->last_nameCasanava (*p).student_idp->student_id910017 *p->last_name+1(*p->last_name))+1D *(p->last_name+2)(p->last_name)[2]s

12. Operator Precedence and Associativity OperatorAssociativity () []. -> ++(postfix) --(postfix)left to right ++(prefix) --(prefix) ! ~ sizeof(type) +(unary) -(unary) &(address) *(dereference) right to left * / %left to right + -left to right >left to right >=left to right == !=left to right & ^ | &&left to right ||left to right ?:right to left = += -= / = %= >>= <<= &= ^= |= right to left,(comma operator)left to right

13. Using structures with functions typedef struct { char name[25]; int employee_id; struct dept deparment; struct home_address *a_ptr; double salary; …. } employee_data;

14. Cont. struct dept { char dept_name[25]; int dept_no; }; e.department.dept_no  (e.department).dept_no employee_data e; e = update(e);

15. Initialization of Structures cardc={13, ’ h ’ }; struct fruitfrt={ “ plum ”, 150); struct home_address { char *street; char *city_and_state; long zip_code; } address = { “ 87 West Street ”, ” Aspen, Colorado ”, 80526}; struct home_address previous_address = {0};

16. Cont. complex[3][3] = { {{1.0, -0.1}, {2.0, 0.2}, {3.0,0.3}} {{4.0,-0.4}, {5.0,0.5}, {60.,0.6}} }; Poker game 설명

17. UNION 구조체와 같은 구문 형식이지만, 각 멤버 들은 같은 기억 장소를 공유 union int_ot_float { int i; float f; }; union int_or_float a,b,c;

18. 예 union int_or_float n; n.i = 4444; n.f = 444.0;  n.i 4444  n.f 0.6227370375e-41 n.f 4444.0  n.i 116672921

19. 다른 예 struct flower { char *name; enum{red, white, blue} color; }; struct fruit { char *name; int calories; }; struct vegetable { char *name; int calories; int cooking_time; }; struct flower_fruit_or_vegetable { struct flowerflw; struct fruitfrt; struct vegetableveg; }; union flower_fruit_or_vegetableffv; ffv.veg.cooking_time = 7;

20. p.425 예 /*In file numbers.x */ typedef union int_or_float { int i; float f; } number; int main() { number n; n.i = 4444; printf( “ i:%10d f:16.10e\n ”, n.i, n.f); n.f = 4444.0; printf( “ i:%10d f:16.10e\n ”, n.i, n.f); return 0; }

21. Bit Field struct pcard { unsigned pips : 4; unsigned suit : 2; } bit_field_number::={int|unsigned} 1 {identi fier} opt :expr expr ::=constant_integral_expression

22. Bit Field 의 예 struct abc { int a : 1, b : 16, c : 16; } x; struct small_integers { unsigned i1:7, i2:7, i3:7, :11, /*align to next word*/ i4:7, i5:7, i6:7; } struct abc { unsigned a:1, :0, b:1, :0, c:1; };

23. p.429 예 /* in file check_bits.x */ #include typedef struct { unsigned b0:8, b1:8, b2:8, b3:8; } word_bytes; typedef struct { unsigned b0:1, b1:1, b2:1, b3:1, b4:1, b6:1, b7:1, b8:1, b9:1, b10:1,b11:1,b12:1, b13:1, b14:1, b15:1, b16:1, b17:1, b18:1, b19:1,b20:1, b21:1,b22:1, b23:1, b24:1, b25:1, b26:1, b27:1, b28:1,b29:1, b30:1,b31: } word_bits; typedef union { intI; word_bitsbit; word_bytesbyte; } word; void main() { word w = {0}; voidbit_print(int)l w.bit.b8 = 1; w.byte.b0 = ‘ a ’ ; printf( “ w.I = %d\n ”, w.I); bit_print(w.I); }

24. ADT Stack ADT(Abstract Data Type) Stack –last-in-first-out(LIFO) –push,pop,top,empty,full,reset 등 Stack 의 구현 (p.431) 문제 – 교재의 Stack 을 사용하여 아래의 식을 계산하라 (13 + 16) * 8 = ??? 단, 위 식은 키보드에서 “ ( ) ” 를 포함한 수식을 입력받으며, 반드시 “ ( ) ” 에 대한 Error 처리가 되어야 한다. –(11 + 3 )) => ERROR!!! 사칙연산이 가능해야 한다