1. Control Structures 4 Control structures control the flow of execution of a program 4 The categories of control structures are: –Sequence –Selection –Repetition 4 A compound statement is used to specify sequential flow

2. Control Structures 4 In sequential flow, control flows from one statement to the next in the compound statement 4 This is the control structure we have been using up until now 4 In this lesson, we will introduce the selection control structures which let us choose between alternative statements

3. Conditions 4 A program chooses among alternative statements by evaluating program variables –age > 65 /* age an int variable */ –This expression is an example of a condition –A condition establishes a criterion for executing or skipping a group of statements –A condition evaluates to either true (represented by 1) or false (represented by 0)

4. Relational & Equality Operators 4 The most common conditions will have one of the following forms: –variable relational-operator variable –variable relational-operator constant –variable equality-operator variable –variable equality-operator constant –The constant may also be the first operand

5. Relational & Equality Operators 4 The relational and equality operators in C are the following: –< less than –> greater than –<= less than or equal to –>= greater than or equal to –== equal to –!= not equal to

6. Logical Operators 4 We can form more complicated conditions using the logical operators && (and), || (or), ! (not) –salary 5 –temperature > 90.0 && humidity > 0.90 –n >= 0 && n <= 100 –0 <= n && n <= 100 –!(0 <= n && n <= 100) 4 Logical expressions are evaluated as are arithmetic expressions

7. The && Operator (and)

8. The || Operator (or)

9. The ! Operator (not)

10. Operator Precedence  We can mix arithmetic and logical operators ( x+5 == y ) thus we must know the precedence of both types of operators –! + - & (unary operators) –* / % –+ - – = > –== != –&& –|| –=

11. Operator Precedence  (x 0.0  x + y < min + max  (x + y) < (min + max)

12. Operator Precedence 4 Let us assume that the following variables have the following values: x 3.0, y 4.0, z 2.0, flag 0. –!flag –x + y /z <= 3.5 –!flag || (y + z >= x - z) –!(flag || (y + z >= x - z))

13. Short-Circuit Evaluation 4 If we have a logical expression with an ‘or’ operator and the first operand evaluates to ‘true’, C doesn’t evaluate the second operand (Why?). This is known as short- circuit evaluation 4 This also works for the ‘and’ operator where the first operand is ‘false’ 4 Why is it important to know this feature of C?

14. Comparing Characters 4 We can also compare characters using the relational and equality operators –‘9’ >= ‘0’ –‘a’ < ‘e’ –‘Z’ == ‘z’ –‘a’ <= ‘A’

15. Logical Assignment 4 The simplest form of a logical expression in C is a single int value or variable –0 represents false –A non-zero value represents true 4 We can use the assignment statement to give a logical value to a variable: –senior_cititzen = 1; –senior_citizen = (age >= 65);

16. Logical Assignment 4 We can now use the variable in a logical expression –senior_citizen && gender == ‘M’ –Evaluates to 1 (true) if senior_citizen is 1 (true) and the character in gender is M

17. Complementing a Condition 4 A logical expression may be complemented by preceding it with the ! Operator 4 A simple condition can be complemented by changing the condition 4 DeMorgan’s theorem tells us that the complement of expr 1 && expr 2 is comp 1 || comp 2, the complement of expr 1 || expr 2 is comp 1 && comp 2

18. Complementing a Condition 4 We can write the complement of –age > 25 && (status == ‘S’ || status == ‘D’) –age <= 25 || (status != ‘S’ && status != ‘D’)

19. The if Statement 4 Now that we are able to write conditions, we will use these conditions to choose among alternative groups of statements 4 The C statement which accomplishes this is the if statement 4 An example of an if statement is if (rest_heart_rate> 56) printf(‘Keep up your exercise program!\n’);

20. The if Statement if (rest_heart_rate> 56) printf(‘Keep up your exercise program!\n’); else printf(‘Your heart is in great health!\n’); 4 This is an example of an if statement with two alternatives –Here, each alternative is a single statement, but in general each alternative can be a compound statement

21. The if Statement 4 We can also write an if statement with a single alternative if (x != 0.0) product = product * x; 4 In this case, if the condition is false, we simply go on to the statement following the if

22. The if Statement 4 If we want to have more than one statement in an alternative, we must use the compound statement construct if (x != 0.0) result = y / x; else { printf(‘You can’t do this!\n’); result = 0.0; }

23. The if Statement if (x != 0.0) result = y / x; else printf(‘You can’t do this!\n’); result = 0.0; 4 What happens in this case? –Indentation has no effect on program execution!

24. The if Statement 4 The general form of the if statement with two alternatives where both alternatives have multiple statements is: if (condition){ true task } else { false task }

25. The if Statement 4 Imagine that we have two variables in a program. One, sex, is a char variable which will have the value ‘m’ or ‘f’. The second, age, is an int variable which holds a person’s age. Write an if statement which examines the variables and prints out ‘Eligible’ for females 63 and older and males 65 and older, and ‘Ineligible’ for everyone else

26. The if Statement 4 We may also need to choose among multiple alternatives (e.g. different prices for men, women, and children). For these cases, we use the nested form of the if statement if (type == ‘m’) price = 25; else if (type == ‘f’) price = 20; else /* type is ‘c’ */ price = 15;

27. The if Statement 4 The general form for the nested if statement is (each statement may be compound): if (condition 1 ) statement 1 else if (condition 2 ) statement 2... else if (condition n ) statement n else statement e