Download
1 / 24
240 likes | 398 Views
13 주 강의. Structures and Unions. Structures. Structure 기본 자료형들로 구성된 새로운 자료형 예 ) struct card { int pips; char suit; };. Structures. struct card { int pips; char suit; }; struct card c1, c2; struct card { int pips; char suit;
E N D
13주 강의 Structures and Unions
Structures • Structure • 기본 자료형들로 구성된 새로운 자료형 • 예) struct card { int pips; char suit; };
Structures struct card { int pips; char suit; }; struct card c1, c2; struct card { int pips; char suit; } c1, c2; ※ pips,suit를 멤버로 가지는 card Type의 c1,c2 선언
값 저장 • c1.pips = 3; • c1.suit = ‘s’; • c2=c1; /* structure 전체를 복사 */ • typedef struct card card; • card c3, c4, c5;
다른 예 • struct fruit { char *name; int calories; } • struct vegetable { char *name; int calories; } • struct fruit a; • struct vegetable b;
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;
Complex number • typedef struct { float re; float im; } complex; • complex a, b, c[100];
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’;
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;
Structures • structure_declaration::=struct_specifier declarator_list; • struct_specifier::=structtag_name struct tag_nameopt {{member_declaration}1+} • tag_name::=identifier • member_declaration::=type_specifier declarator_list; • declarator_list::=declarator{,declarator}0+
Using structures with functions • typedef struct { char name[25]; int employee_id; struct dept deparment; struct home_address *a_ptr; double salary; …. } employee_data;
Cont. • struct dept { char dept_name[25]; int dept_no; }; • e.department.dept_no (e.department).dept_no • employee_data e; e = update(e);
Initialization of Structures • card c={13,’h’}; • struct fruit frt={“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};
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
UNION • 구조체와 같은 구문 형식이지만, 각 멤버들은 같은 기억 장소를 공유 • union int_ot_float { int i; float f; }; union int_or_float a,b,c;
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
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 flower flw; • struct fruit frt; • struct vegetable veg; • }; • union flower_fruit_or_vegetable ffv; • ffv.veg.cooking_time = 7;
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; }
Bit Field • struct pcard { unsigned pips : 4; unsigned suit : 2; } • bit_field_number::={int|unsigned}1{identifier}opt:expr • expr ::=constant_integral_expression
Bit Field의 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; };
p.429 예 /* in file check_bits.x */ #include <stdio.h> 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 { int I; word_bits bit; word_bytes byte; } word; void main() { word w = {0}; void bit_print(int)l w.bit.b8 = 1; w.byte.b0 = ‘a’; printf(“w.I = %d\n”, w.I); bit_print(w.I); }
ADT Stack • ADT(Abstract Data Type) • Stack • last-in-first-out(LIFO) • push,pop,top,empty,full,reset 등 • Stack의 구현(p.431) • 문제 • 교재의 Stack을 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
