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The fundamental data type

The fundamental data type. 제 4 주 강의. Declarations, Expressions, and Assignments. Declaring the type of a variable  set an appropriate amount of space in memory by a compiler  deciding appropriate operations  checking invalid operations and operands (686page 참고 ).

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The fundamental data type

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  1. The fundamental data type 제4주 강의

  2. Declarations, Expressions, and Assignments • Declaring the type of a variable  set an appropriate amount of space in memory by a compiler  deciding appropriate operations  checking invalid operations and operands (686page 참고)

  3. Expressions • Meaningful combinations of constants, variables, operators, and function calls • a+b 5.0*x – tan(9.0/x) • Page 689 참고

  4. Assignments • variable = expr ; • Page 689 참고 • 숙제 ::: a=b=c+d=e  이것이 정상적인 assignment statement임을 문법을 참고로 증명하라!!! 적용된 문법을 보일 것

  5. The fundamental data types • Declaration ::= type identifier {, identifier}0+ ;

  6. Long form • fundamental types • array, pointer, structure ::: derived types

  7. Widely used form • ‘int’ can be omitted

  8. Functionality

  9. 3.3 Characters and the Data Type char • getchar(), putchar() • 8bits (28 =256) • 대문자, 소문자, 숫자, 구두점, 특수문자 • White space characters는 포함 • 제어문자는 제외 • ASCII, EBCDIC

  10. ASCII

  11. Special characters

  12. 출력 • printf(“%c”, ‘\a’) ;  putchar(‘\a’) • printf(“%\”abc\””) ; “abc” • printf(“%cabc%c”, ‘\’’,’\’’) ;  ‘abc’

  13. 출력 예 • int c; int i; • for (i=‘a’; i<=‘z’;++i) printf(“%c”,i); /* abc …z */ • for (c=65; c<=90;++c) printf(“%c”,c); /* ABC … Z */ • for (c=‘0’; c<=‘9’;++c) printf(“%d”,c); /* 48 49 … 57 */ • 01100001(2) == 97 == ‘a’

  14. Integer • 32bits : -231, -231+1 … -1, 0, 1, … … 231-1 • 16bits : -215, -215+1 … -1, 0, 1, … … 215-1 • Nmin_int ≤ i ≤ Nmax_int

  15. Short, long, unsigned • short  16bits • long  32bits • unsigned  32bits 기계: Nmax_unsigned = 232 -1 (일부시스템) 16bits 기계: Nmax_unsigned = 216 –1 (일부시스템) = 65535 •  16bits 기계  32000(short), 33000(long)

  16. 상수표현 방법

  17. The floating types • Float, double, long double

  18. Floating point 값의 표현 • Precision, range • float  0.d1d2…d6 x 10n (-38 ≤ n≤ 38) • double  0.d1d2…d15 x 10n (-308 ≤ n≤ 308) • 123.45123451234512345  0.123451234512345 x 10+3 (유효숫자 15) • long double  long으로 구현

  19. The use of typedef • typedef char uppercase typedef int INCHES, FEET; typedef unsigned long size_t; • uppercase u; INCHES length, width;

  20. The sizeof Operator • sizeof(object) • sizeof(char)  1 • sizeof(char) ≤ sizeof(short) ≤ sizeof(int) ≤ sizeof(long) • sizeof(signed) = sizeof(unsigned) = sizeof(int) • sizeof(float) ≤ sizeof(double) ≤ sizeof(long double)

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