# Fundamentals of Programming Session 7

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1 Fundamentals of Programming Session 7 Instructor: Reza Entezari-Maleki 1 Fall 2014 These slides have been created using Deitel s slides Sharif University of Technology

2 Outlines Formulating Algorithms Case Study 1. Counter-Controlled Repetition 2. Sentinel-Controlled Repetition 3. Nested Control Structures Assignment Operators Increment and Decrement Operators Repetition Essentials for Repetition Statement 2

3 3 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition)

4 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition) 4 The pseudocode algorithm in Fig. 3.7 solves the more general class averaging problem. This algorithm was developed after only two levels of refinement. Sometimes more levels are necessary. Although only integer grades are entered, the averaging calculation is likely to produce a decimal number with a decimal point. The type int cannot represent such a number. The program introduces the data type float to handle numbers with decimal points (called floating-point numbers) and introduces a special operator called a cast operator to handle the averaging calculation.

5 5 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition)

6 6 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition)

7 7 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition)

8 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition) 8 The variable average is defined to be of type float (line 12) to capture the fractional result of our calculation. However, the result of the calculation total / counter is an integer because total and counter are both integer variables. To produce a floating-point calculation with integer values, we must create temporary values that are floating-point numbers. C provides the unary cast operator to accomplish this task. Line 38 average = ( float ) total / counter; includes the cast operator (float), which creates a temporary floating-point copy of its operand, total. The value stored in total is still an integer.

9 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition) 9 Using a cast operator in this manner is called explicit conversion. Most computers can evaluate arithmetic expressions only in which the data types of the operands are identical. To ensure that the operands are of the same type, the compiler performs an operation called promotion (also called implicit conversion) on selected operands. For example, in an expression containing the data types int and float, copies of int operands are made and promoted to float. Cast operators are available for most data types. The cast operator is formed by placing parentheses around a data type name. The cast operator is a unary operator, i.e., an operator that takes only one operand.

10 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition) 10 Figure 3.8 uses the printf conversion specifier %.2f (line 41) to print the value of average. The f specifies that a floating-point value will be printed. The.2 is the precision with which the value will be displayed. If the %f conversion specifier is used (without specifying the precision), the default precision of 6 is used exactly as if the conversion specifier %.6f had been used. When floating-point values are printed with precision, the printed value is rounded to the indicated number of decimal positions, but the value in memory is unaltered. When the following statements are executed, the values 3.45 and 3.4 are printed. printf( "%.2f\n", ); /* prints 3.45 */ printf( "%.1f\n", ); /* prints 3.4 */

11 11 Formulating Algorithms Case Study 2 (Sentinel- Controlled Repetition)

12 Formulating Algorithms Case Study 3 (Nested Control Structures) 12 We ll once again formulate the algorithm using pseudocode and write a corresponding C program. In this case study we ll see the only other structured way control statements may be connected in C, namely through nesting of one control statement within another. Consider the following problem statement: A college offers a course that prepares students for the state licensing exam for real estate brokers. Last year, 10 of the students who completed this course took the licensing examination. Naturally, the college wants to know how well its students did on the exam. You have been asked to write a program to summarize the results. You have been given a list of these 10 students. Next to each name a 1 is written if the student passed the exam and a 2 if the student failed.

13 Formulating Algorithms Case Study 3 (Nested Control Structures) Your program should analyze the results of the exam as follows: Input each test result (i.e., a 1 or a 2). Display the prompting message Enter result each time the program requests another test result. Count the number of test results of each type. Display a summary of the test results indicating the number of students who passed and the number who failed. If more than eight students passed the exam, print the message Bonus to instructor! 13

14 Formulating Algorithms Case Study 3 (Nested Control Structures) 14 After reading the problem statement carefully, we make the following observations: The program must process 10 test results. A counter-controlled loop will be used. Each test result is a number either a 1 or a 2. Each time the program reads a test result, the program must determine if the number is a 1 or a 2. We test for a 1 in our algorithm. If the number is not a 1, we assume that it s a 2. Two counters are used one to count the number of students who passed the exam and one to count the number of students who failed the exam. After the program has processed all the results, it must decide if more than 8 students passed the exam.

15 15 Formulating Algorithms Case Study 3 (Nested Control Structures)

16 16 Formulating Algorithms Case Study 3 (Nested Control Structures)

17 17 Formulating Algorithms Case Study 3 (Nested Control Structures)

18 18 Formulating Algorithms Case Study 3 (Nested Control Structures)

19 19 Formulating Algorithms Case Study 3 (Nested Control Structures)

20 Assignment Operators C provides several assignment operators for abbreviating assignment expressions. For example, the statement c = c + 3; can be abbreviated with the addition assignment operator += as c += 3; Any statement of the form variable = variable operator expression; where operator is one of the binary operators +, -, *, / or % (or others we ll discuss in Chapter 10), can be written in the form variable operator= expression; Thus the assignment c += 3 adds 3 to c. 20

21 Increment and Decrement Operators 21 C also provides the unary increment operator, ++, and the unary decrement operator, --. If a variable c is incremented by 1, the increment operator ++ can be used rather than the expressions c = c + 1 or c += 1. If increment or decrement operators are placed before a variable (i.e., prefixed), they re referred to as the preincrement or predecrement operators, respectively. If increment or decrement operators are placed after a variable (i.e., postfixed), they re referred to as the postincrement or postdecrement operators, respectively.

22 22 Increment and Decrement Operators

23 23 Increment and Decrement Operators

24 Increment and Decrement Operators 24 It s important to note here that when incrementing or decrementing a variable in a statement by itself, the preincrement and postincrement forms have the same effect. It s only when a variable appears in the context of a larger expression that preincrementing and postincrementing have different effects (and similarly for predecrementing and postdecrementing). Of the expressions we ve studied thus far, only a simple variable name may be used as the operand of an increment or decrement operator. Attempting to use the increment or decrement operator on an expression other than a simple variable name is a syntax error e.g. writing ++(x+1).

25 Repetition Essentials Counter-controlled repetition requires: The name of a control variable (or loop counter). The initial value of the control variable. The increment (or decrement) by which the control variable is modified each time through the loop. The condition that tests for the final value of the control variable (i.e., whether looping should continue). 25

26 26 Repetition Essentials

27 27 Repetition Essentials

28 Repetition Essentials 28 You could make the program in Fig. 4.1 more concise by initializing counter to 0 and by replacing the while statement with while ( ++counter <= 10 ) printf( "%d\n", counter ); This code saves a statement because the incrementing is done directly in the while condition before the condition is tested. Also, this code eliminates the need for the braces around the body of the while because the while now contains only one statement. Some programmers feel that this makes the code too cryptic and error prone.

29 for Repetition Statement The for repetition statement handles all the details of counter-controlled repetition. To illustrate its power, let s rewrite the program of Fig The result is shown in Fig

30 30 for Repetition Statement

### In Fig. 3.5 and Fig. 3.7, we include some completely blank lines in the pseudocode for readability. programs into their various phases.

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