1. Control Structures and Data Files

2. Structured Programming
Use simple control structures to organize the solution to a problem
Sequence
Selection
Repetition
yes no
yes no

3. Sequence

4. Selection

5. Repetition

6. Extras Evaluation of alternative solution Error condition
A problem can be solved in many different ways
Which is the best (e.g, faster, memory req)
Error condition
Do not trust user! Check the data. A=b/c;
Be clear about specifications
Generation of Test Data
Test each of the error conditions
Program validation and Verification
Program walkthrough

7. Conditional Expressions
Selection and repetition structures use conditions, so we will first discuss them
A condition is an expression that can be evaluated to be
TRUE (any value > 0) or
FALSE (value of 0)
Conditional Expression is composed of expressions combined with relational and/or logical operators

8. Relational Operators == equality (x == 3) != non equality (y != 0)
< less than (x<y)
> greater than (y>10)
<= less than equal to (x<=0)
>= greater than equal to (x>=y)
!!! a==b vs. a=b !!!

9. Examples a < b x+y >= k/m fabs(denum) < 0.0001 d= b > c
if (d)
a=b+c;
a < b < c ???

10. Logical Operators ! not !(x==0) && and (x>=0) && (x<=10)
|| or (x>0) || (x<0) A B
A && B
A || B !A !B
False
True

11. Examples a<b<c is not the same as (a<b) && (b<c)
a+b * 2 < 5 && 4>=a/2 || t-2 < 10
a=3, b=5, t=4

12. Precedence for Arithmetic, Relational, and Logical Operators

13. Exercise Assume that following variables are declared
a = 5.5 b = k = -3
Are the following true or false
a < k
a + b >= 6.5
k != a-b
!(a == 3*b)
a<10 && a>5
fabs(k)>3 || k<b-a

14. Selection Statements

15. Selection Statements if
if else
switch

16. If statement if(Boolean expression) statement; //single statement
{ //more than one statement
statement1; .
statement n; }

17. Examples if (x>0) { x=sqrt(x); } k++; 4 2 x k -2 2 x k 4 2 x k -2 2

18. if - else statement if (boolean expression) statement1; else{
statement block1 }
statement block2
x=y;
x=y;
k=3;

19. If-else statement
What does the following program do? Assume that x, y, temp are declared.
if (x>y)
temp = x;
else
temp = y; 1 3 ? x y
temp 3 1 ? x y
temp

20. If-else statement Split the following statement into two separate if
if (x>y)
temp = x;
else
temp = y;
if (x>y)
temp = x;
if (x<=y)
temp = y;

21. Exercise
Write an if-else statement to find both the maximum and minimum of two numbers. Assume that x,y, min, max are declared.
if (x>y) Ex: x = 10, y = 5
{ y = 3, y = 4
max = x; x = 6, y = 6
min = y; }
else {
max = y;
min = x;

22. nested if-else if(x > y) if(y < z) k++; if (y<z) k++; else
m++;
j++;
if (y<z)
k++;
else
m++;

23. Exercise int x=9, y=7, z=2, k=0, m=0, j=0; if (x > y) if(y < z)
else
m++;
j++;
What are the values of j, k and m after execution of if statement? 9 7 x y 2 z k m j

24. Exercise int x=3, y=7, z=2, k=0, m=0, j=0; if (x > y) if(y < z)
else
m++;
j++;
What are the values of j, k and m after execution of if statement? 3 7 x y 2 z k m j

25. Exercise What is the output of the following program int a = 5, b = 3;
if (a>10) {
a = 50;
b = 20; }
printf(" a = %d, b = %d\n",a, b);
if (a>10)
a = 50;
b = 20;
printf(" a = %d, b = %d\n",a, b);
if (a>10) {
a = 50;
b = 20; }
printf(" a = %d, b = %d\n",a, b);
if (a>10)
a = 50;
b = 20;
printf(" a = %d, b = %d\n",a, b);

26. Exercise
Given a score and the following grading scale write a program to find the corresponding grade. A B C
0-39 D

27. Solution if ((score >= 80) && (score <=100)) grade = 'A';
else if ((score >= 60) && (score <= 79))
grade = 'B';
else if ((score >= 40) && (score <= 59))
grade = 'C';
else if ((score >= 0) && (score <= 39))
grade = 'D';
else
printf("Invalide Score\n");

28. Alternate Solution printf(“Invalid Score\n”); else if (score >= 80)
grade = 'A';
else if (score >= 60)
grade = 'B';
else if (score >= 40)
grade = 'C';
else if (score >= 0)
grade = 'D';
else
printf("Invalide Score\n");

29. Exercise
What is the value of a at the end of the following if-else statement (chapter3e2.c)
int a = 750;
if (a>0)
if (a >= 1000)
a = 0;
else
if (a <500)
a *= 2;
a *= 10;
a += 3;
if (a>=1000)
a = 0;
else
if (a<500)
a *= 2;
a *= 10;
if (a<500)
a *= 2;
else
a *= 10;

30. Exercise
Write a program that reads 3 numbers a, b and c from user and computes minimum, maximum and median of the numbers.
Example:
a = 2, b = 5, c = 3
minimum = 2, maximum = 5, median = 3
a = 2, b = 2, c = 3
minimum = 2, maximum = 3, median = 2

31. Minimum if ((a<b) && (a<c)) min = a;
else if ((b<a) && (b<c))
min = b;
else if ((c<a) && (c<b))
min = c;

32. Maximum if ((a>b) && (a>c)) max = a;
else if ((b>a) && (b>c))
max = b;
else if ((c>a) && (c>b))
max = c;

33. Median median = b; else if ((c<b) && (b<a))
if ((a<b) && (b<c))
median = b;
else if ((c<b) && (b<a))
else if ((b<a) && (a<c))
median = a;
else if ((c<a) && (a<b))
else if ((b<c) && (c<a))
median = c;
else if ((a<c) && (c<b))

34. Alternate Solution for Median
if (((a<b) && (b<c)) || ((c<b) && (b<a)))
median = b;
else if (((b<a) && (a<c)) || ((c<a) && (a<b)))
median = a;
else if (((b<c) && (c<a)) || ((a<c) && (c<b)))
median = c;

35. Exercise continued Find minimum of 3 numbers
Write an if statement to check if minimum is a
if (condition) what should
min = a; be condition?
Write an if statement to check if minimum is b
Write an if statement to check if minimum is c
Combine 1-3 into a single nested if-else statement
Similarly find maximum of 3 numbers
Find median of 3 numbers

36. Exercise
Write a program that reads a point (x, y) from user and prints the region it resides in.
For example
Enter x, y: 3 -1
Point in Region 4
Enter x, y: -1 -5
Point in Region 3
Region 2
Region 1
Region 3
Region 4

37. Solution if ((x>0) && (y>0)) printf("Region 1\n");
else if ((x<0) && (y>0))
printf("Region 2\n");
else if ((x<0) && (y<0))
printf("Region 3\n");
else if ((x>0) && (y<0))
printf("Region 4\n");

38. Alternate Solution if (x>0) if (y>0) printf("Region1\n"); else

39. Switch Statement switch(expression) { case constant: statement(s);
break;
/* default is optional*/
default: }

40. Switch Statement switch (op_code)
Expression must be of type integer or character
The keyword case must be followed by a constant
break statement is required unless you want all subsequent statements to be executed.
switch (op_code) {
case ‘N’:
printf(“Normal\n”);
break;
case ‘M’:
printf(“Maintenance Needed\n”);
break’
default:
printf(“Error\n”); }

41. Exercise Convert the switch statement into if statement.
switch (op_code) {
case ‘N’:
printf(“Normal\n”);
break;
case ‘M’:
printf(“Maintenance Needed\n”);
default:
printf(“Error\n”); }
if (op_code == ‘N’)
printf(“Normal\n”);
else if (op_code == ‘M’)
printf(“Maintenance Needed\n”);
else
printf(“Error\n”);

42. Exercise
Convert the following nested if/else statements to a switch statement
if (rank==1 || rank==2)
printf("Lower division \n");
else {
if (rank==3 || rank==4)
printf("Upper division \n");
  {
 if (rank==5)
printf("Graduate student \n");
printf("Invalid rank \n"); }

43. Solution { case 1: case 2: printf("Lower Division\n"); break; case 3:
switch(rank) {
case 1:
case 2:
printf("Lower Division\n");
break;
case 3:
case 4:
printf("Upper Division\n");
case 5:
printf("Graduate Student\n");
default:
printf("Invalid Rank"); }

44. Exercise
Write a switch statement that prints the number of days in the month given an integer corresponding to month
int month;
switch (month)
1-12 represents months January – December.
February has 28 days.
April, June, September, November has 30 days
January, March, May, July, August, October, December has 31 days.

45. Solution { case 2: printf("28 Days\n"); break;
switch(month) {
case 2:
printf("28 Days\n");
break;
case 1: case 3: case 5: case 7: case 8: case 10: case 12:
printf("31 days\n");
case 4: case 6: case 9: case 11:
printf("30 days\n");
default:
printf("Invalid month"); }

46. Loop (Repetition) Structures

47. Loop (repetition) statements
while statement
do while statement
for statement

48. while statement while(expression) statement; { . } x=1; while (x<5)
x = x + 1;
while(expression)
statement; { . }
x = 1;
while (x<5) {
x = x+1;
printf(“X=“%d”,x); }

49. Example: lecture2e4.c #include <stdio.h> #define PI 3.141593
int main(void) {
int degrees=0;
double radians;
printf("Degrees to Radians \n");
while (degrees <= 360)
radians = degrees*PI/180;
printf("%6i %9.6f \n",degrees,radians);
degrees += 10; }
return 0;

50. do while do statement; while(expression); { statement1; statement2; .
note - the expression is tested after the statement(s) are executed, so statements are executed at least once.
x=1; do
x = x + 1;
while (x<5);
x = 1; do {
x = x+1;
printf(“X=“%d”,x);
} while (x<5);

51. Example: lecture2e5.c #include <stdio.h> #define PI 3.141593
int main(void) {
int degrees=0;
double radians;
printf("Degrees to Radians \n"); do
radians = degrees*PI/180;
printf("%6i %9.6f \n",degrees,radians);
degrees += 10;
} while (degrees <= 360);
return 0; }

52. for statement for(initialization; test; increment/decrement){ . }
for (x=1; x<5; x++)
printf(“x=%d\n”,x);
for (x=1; x<5; x++) {
printf(“X=“%d\n”,x);
printf(“X^2 = %d\n”,x*x); }

53. for statement initalize test increment/ decrement true statement(s)

54. Examples ? 5 1 3 1 5 7 i sum n fact 1 4 9 int sum =0;
for( i=1 ; i < 7 ; i+=2 )
sum = sum + i;
int fact =1;
for ( n=5;n>1;n--)
fact = fact * n; ? 5 1 3 1 5 7 i
sum n
fact 1 4 9

55. Example: lecture2e6.c #include <stdio.h> #define PI 3.141593
int main(void) {
int degrees;
double radians;
printf("Degrees to Radians \n");
for (degrees=0; degrees<=360; degrees+=10)
radians = degrees*PI/180;
printf("%6i %9.6f \n",degrees,radians); }
return 0;

56. Exercise { statements; } for (count=-2; count<=5; count++)
Determine the number of times that each of the following for loops are executed.
for (k=3; k<=10; k++) {
statements; }
for (count=-2; count<=5; count++)

57. Exercise Write a loop to print Hello 10 times
for (i=1; i<=10; i++) {
printf(“Hello\n”); }
Change it to print Hello 20 times
Change it to print the count besides each hello
Hello 1
Hello 2

58. Exercise Write program to compute the following using a for loop
Result = 55
Result = 45

59. Exercise Convert the following for loop to while loop
for (i=5; i<10; i++) {
printf(“ i = %d \n”, i); }
i=5;
while (i<10)
i++;

60. Exercise
Write a loop to compute xy for y integer without using pow(x,y)
res=1;
for (i=1; i<=y; i++) {
res = res * x; }
20 = 1
21 = 2
22 = 2*2
23 = 2*2*2

61. Exercise Write a program to compute the following total=0; sofarx=1;
for (i=0; i<=m; i++) {
total = total +sofarx;
sofarx = sofarx * x; }
total=0;
for (i=0; i<=m; i++)
total = total + pow(x, i);

62. Exercise Write a program to compute the following ln2=0;
for (i=1; i<=n; i++)
if ( i % 2 == 0)
ln2 = ln / i;
else
ln2 = ln / i;

63. Exercise What is the output of the following program?
for (i=1; i<=5; i++) {
for (j=1; j<=4; j++)
printf(“*”);
printf(“\n”); }
Available on class webpage: lecture2e7.c
Output
****

64. Exercise What is the output of the following program?
for (i=1; i<=5; i++) {
for (j=1; j<=i; j++)
printf(“*”);
printf(“\n”); }
Available on class webpage: lecture2e8.c
Output * **
***
****
*****

65. Exercise Modify the following program to produce the output.
for (i=A; i<=B; i++) {
for (j=C; j<=D; j++)
printf(“*”);
printf(“\n”); }
Output
*****
****
*** ** *

66. Exercise Write a program using loop statements to produce the output.* **
***
****
*****
for (i=1; i<=5; i++) {
for (j=1; j<=5-i; j++)
printf(" ");
for (k=1; k<=i; k++)
printf("*");
printf("\n"); }

67. break statement break; terminates loop
execution continues with the first statement following the loop
sum = 0;
for (k=1; k<=5; k++) {
scanf(“%lf”,&x);
if (x > 10.0)
sum +=x; }
printf(“Sum = %f \n”,sum);
Available on class webpage: lecture2e9.c

68. continue statement continue;
forces next iteration of the loop, skipping any remaining statements in the loop
sum = 0;
for (k=1; k<=5; k++) {
scanf(“%lf”,&x);
if (x > 10.0)
sum +=x; }
printf(“Sum = %f \n”,sum);
Available on class webpage: lecture2e10.c