2. While Loop Structure while (condition) { …. // This line will process when while condition is true }

3. do - While Loop Structure do { …. // This line will process first then check while condition } while (condition); // if while condition is true then process inside do again

4. For Loop Structure for (initial; condition; increment) { …. // This line will process when for condition is true } initial = initialize value of index (do at once) Condition = check whether true or false Increment = after process then define the increment or decrement of index

5. Nested For Loop Structure for (initial; condition; increment) { …. (outer loop line) for (initial; condition; increment) { …. (inner loop line) } ….. (outer loop line) }

6. How to use “FOR” instead of “WHILE” in the following example main() { int i; i =1; while(i <= 10) { printf(“%d”,i); i = i+1; }

7. How to use “FOR” instead of “WHILE” in the following example main() { int i; i =1; while(i <= 10) { printf(“%d”,i); i =i+1; } main() { int i; for (i=1;i <= 10;i=i+1) { printf(“%d”,i); }

8. A1 : Given The Following Code

9. i = 0 ; max = 6 ; Solution: Loop1 : i=0, max=6  0 Loop2 : i=3, max=6  3 Loop3 : i=6, max=6 

10. A1 : Given The Following Code i = 0 ; max = 6 ; Result: 03

11. A1 : Given The Following Code i = -3 ; max = 6 ; Solution: Loop1 : i=-3, max=6  -3 Loop2 : i=0, max=6  0 Loop3 : i=3, max=6  3 Loop4 : i=6, max=6 

12. A1 : Given The Following Code i = -3 ; max = 6 ; Result: -303

13. A1 : Given The Following Code i = 2 ; max = ++i +i +3 ; Solution: Loop1 : i=3, max=9  3 Loop2 : i=6, max=9  6 Loop3 : i=9, max=9  ++i + i +3 = 3+3+3  9

14. A1 : Given The Following Code i = 2 ; max = ++i +i +3 ; Result: 36

15. A1 : Given The Following Code max = 8 ; i = max/2; Solution: Loop1 : i=4, max=8  4 Loop2 : i=7, max=8  7 Loop3 : i=10, max=8 

16. A1 : Given The Following Code max = 8 ; i = max/2; Result: 47

17. A2 : What is the output of the following program?

18. Result: x is 3(n=7) Solution: Loop1 : x=15, n=4  13 Loop2 : x=13, n=4  11 Loop3 : x=11, n=4  9 Loop4 : x=9, n=4  7 Loop5 : x=7, n=4  5 Loop6 : x=5, n=4  3 Loop7 : x=3, n=4 

19. B9 : What is the output of the following program?

21. #include int main(void) { int x; for(x=3;x<20;x++) { if(x==5) continue; if(x==10) break; printf("%d ",x); } return 0; } Result: b. 3 4 6 7 8 9 Output Loop1(x=3)  3 Loop2(x=4)  4 Loop3(x=5)  continue; Loop4(x=6)  6 Loop5(x=7)  7 Loop6(x=8)  8 Loop7(x=9)  9 Loop3(x=10)  Break;

22. B11 : In the following C program, how many times will i be printed out on the screen?

23. B11 : In the following C program, how many times will i be printed out on the screen? #include int main(void) { int i=10; do{ printf("i"); i++; }while(i<10); } Result: d. 1 Output Loop1(i=10)  i Loop2(i=11) 

25. Array Structure int digits[10] = {1,2,3,4,5,6,7,8,9,10}; static float x[6] = { 0,0.25,0,0.50,0,0}; char color[] = {'R','E','D'}; char color2[] = "RED" digits[0] = 1 x[0] = 0 color[0] = 'R‘ digits[1] = 2 x[1] = 0.25 color[1] = 'E' digits[2] = 3 x[2] = 0 color[2] = 'D' digits[3] = 4 x[3] = 0.50 digits[4] = 5 x[4] = 0 color2[0] = 'R' digits[5] = 6 x[5] = 0 color2[1] = 'E' digits[6] = 7 color2[2] = 'D' digits[7] = 8 color2[3] = '\0' digits[8] = 9 digits[9] = 10

27. Link to Bubble Sort Example

28. B5:

29. Solution: a.int a[3]={2,3,4};  b.int a[] = {2,3,4};  c.int a[5] = {2,3,4};  d.int a[3] = {2,3,4,0,0};  234 23400 234 23400 234

30. B6: 5 Rows 4 Columns

31. B6: Solution: a.int array[10][4];  array with 10 rows and 4 columns b.int array[4][10];  array with 4 rows and 10 columns c.int array[4,10];  ERROR d.int a {10} {4};  ERROR

32. B18:

33. B10:

36. String is defined as an ARRAY of CHARACTERS String is terminated by a special character which is null parameter (\0). We declare a string by char color[] = “blue”; char color[] = {‘b’, ‘l’, ‘u’, ‘e’,‘\0’}; char msg[10] = “Computer”; char msg[10] = {‘C’,’o’,’m’,’p’,’u’,’t’,’e’,’r’,’\0’};

37. String Input Functions scanf (“%s”,name);  Scan until first space gets(name);  Scan until end of line sscanf(data,”%s”,name);  Read from data and keep string in name EXAMPLE: This is a Book scanf : This gets: This is a Book sscanf : This

38. String Output Functions #include main() { int i; char name[]= “Smith”; printf(“Name = %s”, name); puts(name); for(i=0;i<5;i++) { printf(“%c”, name[i]); }

39. Link to Example of ctype.h

40. string.h strcat(); Appends the string pointed to by str2 to the end of the string pointed to by str1. strncat(); Appends the string pointed to by str2 to the end of the string pointed to by str1 up to n characters long. strchr(); Searches for the first occurrence of the character c (an unsigned char) in the string pointed to by the argument str. strcmp(); Compares the string pointed to by str1 to the string pointed to by str2. strncmp(); Compares at most the first n bytes of str1 and str2. Stops comparing after the null character. strcpy(); Copies the string pointed to by str2 to str1. strncpy(); Copies up to n characters from the string pointed to by str2 to str1.

41. string.h strcspn(); Finds the first sequence of characters in the string str1 that does not contain any character specified in str2. strlen(); Computes the length of the string str up to but not including the terminating null character. strpbrk(); Finds the first character in the string str1 that matches any character specified in str2. strrchr(); Searches for the last occurrence of the character c (an unsigned char) in the string pointed to by the argument str. strspn(); Finds the first sequence of characters in the string str1 that contains any character specified in str2. strstr(); Finds the first occurrence of the entire string str2 (not including the terminating null character) which appears in the string str1. strtok(); Breaks string str1 into a series of tokens. strxfrm(); Transforms the string str2 and places the result into str1.

43. B13:

45. B21:

46. Solution printf("%d %c %c\n", toupper('a')+1, toupper('b')+1, tolower('C')-2); toupper('a')+1  A (65) +1 = 66  %d  66 toupper('b')+1  B(66) +1 = 67  %c  C tolower('C')-2);  c(99) – 2 = 97  %c  a

47. B14:

49. B22:

50. Solution a. string.h  This header file defines several functions to manipulate C strings and arrays b. stdio.h  This header file for the standard functions that deal with input and output c. stdlib.h  This header defines several general purpose functions, including dynamic memory management and random d. ctype.h  This header allow the programmer to check, compare, and manipulate characters

51. A5:

52. Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.1 Convert the character stored in variable c to an uppercase letter. Assign the result to variable c.

53. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.1 Convert the character stored in variable c to an uppercase letter. Assign the result to variable c. #include main() { int c = 'a'; c = toupper(c); printf("%c",c); } Answer

54. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.2 Convert the string “8.635” to double and print the value.

55. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.2 Convert the string “8.635” to double and print the value. #include main() { char s2[100] = "8.635"; printf("%.3f",atof(s2)); } Answer

56. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.3 Copy the string stored in array s2 into array s1.

57. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.3 Copy the string stored in array s2 into array s1. #include main() { char s1[100], s2[100] = "8.635"; strcpy (s1,s2); printf("%s",s1); } Answer

58. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.4 Determine the length of the string in s1. Print the result.

59. A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char. 5.4 Determine the length of the string in s1. Print the result. #include main() { char s1[100] = "8.635"; printf("%d",strlen(s1)); } Answer

61. Why? Main reason – Often be used more than once in a program – Fit naturally with a top-down design approach – Provide a natural method for dividing a programming task among team – Can be tested individually

62. Example: No Output void print_hi() { printf(“Hi”); } No Output

63. Example: No Input void print_hi() { printf(“Hi”); } void print_hi(void) { printf(“Hi”); } No Input

64. Example: Input as Integer int factorial(int n) { int ans=1; int i; for (i=2;i<n;i++) { ans*=i; } return ans; } n is input

65. Example: Output as Integer int factorial(int n) { int ans=1; int i; for (i=2;i<n;i++) { ans*=i; } return ans; } Output as Integer Output value of ans

66. Example: Two Input void print_chars(char ch, int n) { int i; for (i=0;i<n;i++) { printf(“%c”,ch); }

67. Function Prototypes double cal_area(double r); void main() { … area= cal_area(r); … } double cal_area(double r) { return 3.14*r*r; } double cal_area(double r) { return 3.14*r*r; } void main() { … area= cal_area(r); … }

68. Global vs Local Variable int pi=3.14; double cal_area(double r) { return pi*r*r; } void main() { … area= cal_area(r); cir= 2*pi*r; } double cal_area(double r) { int pi=3.14; return pi*r*r; } void main() {… int pi=3.14; area= cal_area(r); cir = 2*pi*r; }

69. Extract function from code void main() { double r,area; printf(“input radius>”); scanf(“%lf”,&r); area= 3.14*r*r; printf(“area is %lf”,area); } r is an input Type of r is double Need to output as double Link to Example of Function

70. B4:

71. Global Variables Local Variables

72. B4: Solution: a.local variable;  A variable defined inside a function b.general variable  A variable as input to the function c.prototype variable  A function prototype is a function header without implementation. d. global variable  A variable defined outside a function

73. B7:

79. What is the output? #include int x,y,z; void junk(int x,int y,int z) { x++; y++; z++; printf(“%10d%10d%10d\n”,x,y,z); } main() { x=1; y=2; z=3; printf(“%10d%10d%10d\n”,x,y,z); junk(x,y,z); printf(“%10d%10d%10d\n”,x,y,z); }

80. What is the output? #include int x,y,z; void junk(int x,int y,int z) { x++; y++; z++; printf(“%10d%10d%10d\n”,x,y,z); } main() { x=1; y=2; z=3; printf(“%10d%10d%10d\n”,x,y,z); junk(x,y,z); printf(“%10d%10d%10d\n”,x,y,z); } 1 2 3 2 3 4 1 2 3

82. What is the output? #include int x,y,z; void junk(int *x,int *y,int *z) { (*x)++; (*y)++; (*z)++; printf(“%10d%10d%10d\n”,*x,*y,*z); } main() { x=1; y=2; z=3; printf(“%10d%10d%10d\n”,x,y,z); junk(&x,&y,&z); printf(“%10d%10d%10d\n”,x,y,z); }

83. What is the output? #include int x,y,z; void junk(int *x,int *y,int *z) { (*x)++; (*y)++; (*z)++; printf(“%10d%10d%10d\n”,*x,*y,*z); } main() { x=1; y=2; z=3; printf(“%10d%10d%10d\n”,x,y,z); junk(&x,&y,&z); printf(“%10d%10d%10d\n”,x,y,z); } 1 2 3 2 3 4

84. Call by Reference We can solved problem of multiple output from function using Call by Reference Link to Example of Call by Reference

85. B1:

86. Solution: a.int x;  creates a variable x as integer type b.ptr x;  Not found this type of declaration c.int &x;  creates a reference to an int named 'x' d.int *x;  x is a pointer to an integer

87. B2:

88. Solution: a.*a;  value of memory address of variable named 'a' b.a;  variable named 'a c.&a;  memory address of variable named 'a' d.address(a);  function named 'address'

89. B15:

91. B16:

92. Solution: a.char *  Example: “123”, “123abc”, ”abc” b.float *;  Example: 1.23, 1.0, 100 c.int  Example: 1, 2, 3, 10, 100, 1000 d.char  Example: ‘a’, ‘b’, ‘1’, ‘2’, ‘3’

93. B17:

94. Solution: a.char *s=“Test”  b.char s[5]={'t','e','s','t',0};  c.char s[]=“Hello”  d.char s[5]=“Hello”  TEST\0 TEST Hello Hello!!@

95. B20:

96. 2 p 3451 a. p=a+3; 2 p 3451 c. p++; 23451 d. *q=(*p); P q

97. B3:

98. Solution: a.*p;  value of the variable pointed to by pointer p b.p;  memory address of the variable pointed to by pointer p c.&p;  memory address of pointer p d.address(p);  function named 'address'

99. B12:

101. A4 : What is the output of the following program? Link to Solution

102. A4 : What is the output of the following program? Result: x = 32 y = 20

104. Structure Structures are the C equivalent of records. A structure type can define in 2 ways, which are struct Foo { int x; int array[100]; }; Usage: struct Foo f; f.x = 54; f.array[3]=9; typedef struct { int x; int array[100]; } Foo; Usage: Foo f; f.x = 54; f.array[3]=9; Link to Example of Structure

105. Structure Passing Structs as Parameters Structs are passed by value, just like primitives void func(struct Foo foo){ foo.x = 56; foo.array[3]=55; } void main(){ struct Foo f; f.x = 54; f.array[3]=9; func(f); printf("%d and %d",f.x,f.array[3]); }

106. Structure Passing Structs as Parameters To have changes occur, send a pointer to the struct void func(struct Foo* foo){ (*foo).x = 56; (*foo).array[3]=55; } void main(){ struct Foo f; f.x = 54; f.array[3]=9; func(&f); printf("%d and %d",f.x,f.array[3]); }

107. What will be the output of the following code and why? #include typedef struct { char *a, *b, *c; } colors; void funct(colors sample){ sample.a = “Cyan”; sample.b = “Brown”; sample.c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } main() { colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); funct(sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c); }

108. What will be the output of the following code and why? #include typedef struct { char *a, *b, *c; } colors; void funct(colors sample){ sample.a = “Cyan”; sample.b = “Brown”; sample.c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } main() { colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); funct(sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } red green blue Cyan Brown Yellow red green blue

109. If we want output as followed, how to change the program? #include typedef struct { char *a, *b, *c; } colors; void funct(colors sample){ sample.a = “Cyan”; sample.b = “Brown”; sample.c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } main() { colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); funct(sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } red green blue Cyan Brown Yellow

110. If we want output as followed, how to change the program? #include typedef struct { char *a, *b, *c; } colors; void funct(colors *sample){ sample.a = “Cyan”; sample.b = “Brown”; sample.c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } main() { colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); funct(sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } red green blue Cyan Brown Yellow

111. If we want output as followed, how to change the program? #include typedef struct { char *a, *b, *c; } colors; void funct(colors *sample){ sample.a = “Cyan”; sample.b = “Brown”; sample.c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } main() { colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); funct(&sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } red green blue Cyan Brown Yellow

112. If we want output as followed, how to change the program? #include typedef struct { char *a, *b, *c; } colors; void funct(colors *sample){ (*sample).a = “Cyan”; (*sample).b = “Brown”; (*sample).c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } main() { colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c); funct(&sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c); } red green blue Cyan Brown Yellow

113. B27:

114. B8: C++ Spring 2000 Arrays typedef struct tnode { int a,b,c,d;} node; main() { node x; int *msg = &x;} 2 msg 345 x.a x.b x.c x.d

115. B8: C++ Spring 2000 Arrays typedef struct tnode { int a,b,c,d;} node; main() { node x; int *msg = &x;} 2 msg 345 x.a x.b x.c x.d 4 Bytes

116. Use the given part of the program to answer questions 28-30

117. B28:

119. B29: 4 Byte 40 Byte 118 Byte 4 Byte

120. B29: 4 Byte 40 Byte 118 Byte 4 Byte

121. B30:

122. Wrong type argument NULL undeclared

123. A3 : What is the output of the following program? Link to Solution

124. A3 : What is the output of the following program? Result: 8 6 6

129. B19: Solution: a.fseek  sets the file position indicator for the stream pointed to by stream. b.fscanf  read data from file into specified variable. c.rewind  Moving the file position marker back to the beginning of the file. d.feof  Check whether the end-of-file marker is reached.

130. B19: Solution: a.fseek  sets the file position indicator for the stream pointed to by stream. b.fscanf  read data from file into specified variable. c.rewind  Moving the file position marker back to the beginning of the file. d.feof  Check whether the end-of-file marker is reached.

131. B23:

134. B24:

136. B25:

138. B26: