printf int number 10 char response y double

표준 출력 함수 printf int number = 10; char response = ‘y’; double root 1 = 3. 25; char str[21] = “Normal termination”; printf(“%d”, 15); 15 printf(“%d”, number); 10 printf(“%d”, number + 5); 15 printf(“Enter value for a: ”); Enter value for a: printf(“%s”, “Enter value for a: “); Enter value for a: printf(“%c”, ‘a’); a printf(“%c”, response); y printf(“%f”, 1. 25); 1. 250000 printf(“%f”, 1. 2500); 1. 250000 printf(“%f”, -1. 0000); -1. 000000 printf(“%f”, 1. 2500 e 1); 12. 500000 printf(“%f”, root 1); 3. 250000 printf(“%lf”, root 1); 3. 250000 printf(“%e”, root 1); 3. 250000 e+00 printf(“%s”, str); Normal termination Chapter 7 문제해결을 위한 C언어 프로그래밍 2

출력 서식화(formatting) l 예제 7. 2 : int number가 20이라면 » printf(“%10 d”, number); � � � � 20 l 예제 7. 3 » printf(“%-10 d”, number); 20 � � � � l 예제 7. 4 » printf(“%010 d”, number); 000020 l 예제 7. 5 : int number가 35라면 » printf(“%+10 d”, number); � � � � +35 l 예제 7. 6 : double number가 34. 5678이라면 » printf(“%15. 3 f”, number); � � � � 34. 568 l 예제 7. 7 » printf(“%15 s”, “Greetings!”); � � � Greetings! l 예제 7. 8 : double number가 155. 56788이라면 » printf(“%15. 3 f”, number); � � � � 155. 568 » printf(“%15. 0 f”, number); � � � 156 » printf(“%15. 3 e”, number); � � � 1. 556 e+02 l 예제 7. 9 » printf(“%+015. 3 f”, number); � � � � 155. 568 Chapter 7 문제해결을 위한 C언어 프로그래밍 3