Control structures the basic operations of a c c programming language are specified using the follow

Control Structures The basic operations of a C/C++ programming language are specified using the following statements or structures FOR MORE CLASSES VISIT www.tutorialoutlet.com

Control Structures The basic operations of a C/C++ programming language are specified using the following statements or structures: the input statement the output statement the assignment statements the single-way selection structure the two-way selection structure the multiple-way selection structure the counter-controlled iteration structure, and the logically-controlled iteration structure. The single-way selection, the two-way selection, the multiple-way selection, the countercontrolled iteration and the logically-controlled iteration are collectively called control structures. The specification of a controlled structure depends on a logical expression called condition. Controlled structures may also be specified using the following flowchart symbols: Flowchart Symbols

SYMBOL (a) (b) (c) (d) NAME Terminal Input/Output Process Flow Lines DESCRIPTION Indicates the beginning or the end of an algorithm.

Indicates an input or an output operation.

Indicates a computation or a data manipulation. Used to connect the symbols and to indicate the logic flow.

Indicates a decision point in an algorithm. (e) Decision (f) Connector Š 2013 Gilbert Ndjatou Indicates an entry or an exit from another part of the flowchart. Page 57 Logical Expressions A logical expression (or condition) is an expression that evaluates to either true or false. There are two types of conditions: simple conditions and Compound conditions. Simple Conditions A simple condition has the following syntax: <arithmetic expression> <relational operator> <arithmetic expression> Relational Operators C/C++ Symbol Meaning < is less than > is greater than == is equal to <= is less than or equal to >= is greater than or equal to != is not equal to Evaluations of simple conditions Assuming that the variables are defined and initialized as follows: int num1 = 5 , num2 = 7 , num3 = 2;

float fnum = 11.75; char ch = ‘K’; Evaluate the following conditions: a) num1 >= 5 b) fnum + 7.2 != fnum / 2 c) fnum * 2 > num1 + 20 © 2013 Gilbert Ndjatou d) num2 + 3 == num3 * 4 e) 3 * num1 + 4 < num3 * 2 f) ‘A’ <= ch Page 58 Solutions a) num1 >= 5 5 >= 5 True d) num2 + 3 == num3 * 4 7 + 3 == 2 * 4 10 == 8 False b) fnum + 7.2 != fnum / 2 11.75 + 7.2 != 11.75 / 2 18.95 != 5.875 True e) 3 * num1 + 4 < num3 * 2 3*5+4 < 2*2 19 < 4 False c) fnum * 2 > num1 +20 11.75 *2 > 5 + 20

13.75 > 25 False f) ‘A’ <= ch ‘A’ <= ‘K’ True Characters are ordered according to their ASCII code representations: ‘0' < ‘1' < ‘2' < . . . < ‘9' < ‘A’ < . . . < ‘Z’ < ‘a’ . . . < ‘z’. Relational operators have a lower precedence than arithmetic operators: o In the absence of parentheses, arithmetic operations are evaluated before any relational operation is evaluated. However, it is a good programming practice to use parentheses to clarify the meaning of logical expressions.

For example: (fnum * 2) > (num1 + 20) Compound Conditions A compound condition is built from simple conditions using logical operators.

Logical Operators C++ Symbol Meaning Evaluation AND <condition1> && <condition2> is true if and only if both conditions are true || OR <condition1> || <condition2> is true if and only if at least one of the two conditions is true ! NOT && © 2013 Gilbert Ndjatou !<condition> is true if and only if <condition> is false Page 59 Evaluations of compound conditions True || True True || False True False || False True False True True && True True && False True False || True False && True False False && False False False There

is a short-circuit evaluation of a compound condition when you do not have to evaluate the right-most condition to get the true value of the compound condition: True || True and True || False True Therefore True True || True Also, False && True and False && False False Therefore False False && False Exercise CS1* Assuming that the variables are defined and initialized as follows: int num1 = 5, num2 = 7, num3 = 2; float fnum = 11.75; char ch = ‘K’; Compute the true value of each of the following compound conditions: a) num1 >= 5 || num2 + 3 == num3 b) ‘A’ <= ch && fnum + 7.2 != 2.5 c) !(3 * num1 + 4 < num3 * 2) d) num2 + 3 == num3 * 4 || ch >= ‘Z’ e) num1 < 3 && num2 > num3 f) !(fnum * 2 < num1 + 20) Solution a) num1 >= 5 || num2 + 3 == num3 5 >= 5 || True True © 2013 Gilbert Ndjatou d) num2 + 3 == num3 * 4 || ch >= ‘Z’ 7 + 3 == 2 * 4 || ‘K’ >= ‘Z’

10 == 8 || False False || False False Page 60 b) ‘A’ <= ch ‘A’ <= ‘K’ True True && fnum + 7.2 != 2.5 && 11.75 + 7.2 != 2.5 && 18.95 != 2.5 && True True e) num1 < 3 && num2 > num3 5 < 3 && False False c) !(3 * num1 + 4 < num3 * 2) !( 3 * 5 + 4 < 2 * 2) !( 19 < 4) ! False True f) !(fnum * 2 < num1 + 20) !(11.75 * 2 < 5 + 20) !( 23.50 < 25) !True

False Precedence of C/C++ Operators Operator Order of Evaluation Precedence !

Unary – right to left 7 * / % + < <= > <= == != && || left to right 6 left to right 5 left to right 4 left to right left to right left to right 3 2 1 In the evaluation of an expression, operators with higher precedence are evaluated before those with

lower precedence. Operators with the same precedence are evaluated in the order specified in the “order of evaluation” column. © 2013 Gilbert Ndjatou Page 61 Exercise CS2 Assuming that the variables are defined and initialized as follows: int num1 = 9 , num2 = 5 , num3 = 10; float fnum = 12.50; char ch = ‘P’; Evaluate the following conditions (using short circuit evaluation whenever possible): a) 2 * num1 - 5 >= 9 || fnum / 2 + 10 <= 6.5 b) num1 + num2 == num3 + 5 && 2 * num3 <= 4 * num2 c) ! (num1 + 5 <= 13) d) num1 + 5 = = 24 || num1 – 1 > num2 – 5 e) 2 * num1 - 5 >= 9 && fnum / 2 + 10 <= 6.5 f) num1 - num2 <= num1 / 3 || num1 * num2 > 100 g) num1 - 5 >= num3 || num2 < 15 && num1 >= 9 Accuracy of Floating-Point Values The fact that floating-point values are approximated inside a computer makes it difficult to test for the equality of floating-point values.

For example, if the variable fnum is defined as follows: float Then, the condition: fnum = 7.1; fnum / 2 == 3.55 may not be true.

The problem may be solved by assuming that two floating-point values are equal if their difference

is relatively very small. This is done by testing if the absolute value of their difference is less than a certain value chosen by the programmer (for example 0.000001). Using the library function fabs() that takes as argument a floating-point value and returns its absolute value, the condition: value1 == value2 is replaced with the condition: fabs(value1 - value2) < 0.000001 which tests whether the difference between the two values is small enough so that we can make them equal. Š 2013 Gilbert Ndjatou Page 62 Evaluation of True and False In the C/C++ programming language, false is represented by the integer value 0, and any value other than 0 represents true. Therefore, the following two conditions are equivalent (have the same true value): o Conditions value != 0 and value o Conditions value == 0 and ! value (value is not zero means that it is true) (value is zero means that !value is true) bool Data Type A variable with data type bool can only hold the bool values true and false.

Examples: bool flag = true; bool condValue = false, HighTemperature, ExtremeTemperature; double temperature; HighTemperature = false cin >> temperature; HighTemperature = (temperature >= 120)

ExtremeTemperature = (HighTemperature || (temperature <= -20); Exercise CS3 1. a) b) c) Which of the following conditions are equivalent (that means have the same true value)? num != 0 d) !num num == 0 e) !(num == 0) num 2. Suppose that a person’s age is stored in the variable age, his number of children in the variable numChild, his salary in the variable salary, and his height in the variable height. Write the relational expressions to specify the following conditions: a. he is 45 year old. b. he is more than 5.7 feet tall. c. his salary is between 35,000 and 50,000. d. he does not have 3 children. e. he is either 6.0 feet tall, or he has less than 4 children. f. he is not older than 35 and he has 2 or 3 children Š 2013 Gilbert Ndjatou Page 63 Two-Way Selection using the if-else Structure The solutions of some problems require the CPU to test a condition and to select the action to be performed based on whether or not that condition is true. Example:

Suppose that at a party, you serve alcohol drinks only to guests who are 21 or older, and a juice to guests who are less than 21.

When a guest comes to get a drink, you first test his age: if he is 21 or older, you serve him an alcohol drink; otherwise you serve him a juice. The solutions of these types of problems are written using a two-way selection structure which is specified using a flowchart as follows: False Condition True TStatement F-Statement Next-Statement It says to do the following: Test condition a. if it is true, perform the actions (T-Action) specified by TStatement, b. otherwise, perform the actions (F-Action) specified by F-Statement.

Next-Statement is the statement that says what to do after any of the actions (F-Action or T-Action) above has been executed. A two-way selection structure is specified in C++ using the if - else structure as follows: if (<condition>) <T-statement> else <F-statement> <Next-Statement> Š 2013 Gilbert Ndjatou Page 64 <T-statement> is a single statement that specifies the T-Action <F-statement> is a single statement that specifies the <FAction>. Case Study CS1 Problem Statement

Write a program to read the age of an individual, and to output the message “serve alcohol drink” if he is 21 or older, and the message “serve juice” otherwise. Your program will then output the message “thank you for using this program.” Program Logic output: “Serve alcohol” or “Serve Juice”, depending on the individual’s age. Input: an individual’s age. Variables: age (int) to hold an individual’s age. Algorithm Specification (using a Flowchart) start Read age False True age >= 21 Write “serve a juice” Write “serve alcohol drink” Write “Thank you for using this program” Stop © 2013 Gilbert Ndjatou Page 65 Figure CS1 Using the if-else Structure /******************************************************* ****************** Program to read the age of an individual and to output the type of drink that he should be served ******************************************************** *****************/ #include <iostream> using namespace std;

#define DRINKAGE 21 int main() { int age; // to hold an individual’s age /*----------------- read the individual’s age------------------------*/ cout << “\n\nEnter the individual’s age please:\t”; cin >> age; /*--------determine what type of drink he should be served ---------*/ if (age >= DRINKAGE) // he is over the drinking age cout << endl << “Serve alcohol drink”; else // he can not drink alcohol cout << endl << “Serve juice”; cout << endl << “Thank you for using this program”; return (0); } Case Study CS2 Problem Statement Write a program to read a positive integer value and to determine if it is even or odd. If it is even, print it with the message “EVEN”; otherwise, print it with the message “ODD”. Program Logic

Output: the input value with the message “EVEN” or “ODD”, depending on whether the input value is even or odd. Input: an integer value. Variable: num (int) to hold the input value. Note: An integer value is even if the remainder in its division by 2 is 0; otherwise, it is odd. © 2013 Gilbert Ndjatou Page 66 Algorithm Specification (using a Flowchart) Start Read num False True num % 2 = 0 Write num, “even” Write num, “odd” Stop Figure CS2 Using the if - else Structure /******************************************************* ***************** Program to read an integer value and to determine if it is even or odd ******************************************************** ****************/ #include <iostream> using namespace std; int main() { int num; // to hold the value read /*---------------- read in an integer value--------------------------*/ cout << “\n\nEnter an integer value please:\t”; cin >> num; /*----------------determine if it is even or odd --------------------*/

if (num %2 == 0) // it is even cout << endl << num << “\tEVEN”; else // it is odd cout << endl << num << “\tODD”; return ( 0 ); } Exercise CS4 Write a C++ code segment to read an integer value and to determine if it is a multiple of 5. If it is a multiple of 5, print it with the message “MULTIPLE OF 5”, otherwise, print it with the message NOT MULTIPLE OF 5.” © 2013 Gilbert Ndjatou Page 67 Exercise CS5 Write a C++ code segment to read a character and to determine if it is a letter of the alphabet. If it is a letter of the alphabet, print it with the message, “it is a letter” otherwise, print it with the message, “it is

not a letter.” Compound Statement A compound statement has the following syntax: { (One or more statements to be executed as a block) } Example { num2 = num1 + 5; num1 = num1 +1; cout << endl << “num1=\t” << num1 << “\n num2=\t” << num2; } The statements of a compound statement are executed one after another, starting with the first one in the sequence. A compound statement may appear anywhere in a C++ program where a single statement may appear. If you have to specify two or more statements as the Tstatement or the F-statement of an if-else structure, these statements must be specified as a compound statement. Case study CS3 Problem Statement Write a program to read an integer value and to do the following: If the value read is zero, print it with the message “INVALID DIVISOR”. Otherwise read a second value and then compute and print the quotient and the remainder in the division of the second value by the first. At the end write the message “Thank you for using this program.” © 2013 Gilbert Ndjatou Page 68 Program Logic Output: the quotient and the remainder in the division of the second value by the first or the

message “0 is invalid divisor” depending on the input. Input: one or two integer values, depending on the value of the first. Variables: divisor (int) to hold the first value. dividend (int) to hold the second value. Note: The second value is input and the quotient and the remainder in the division of the second value by the first are computed only if the first value is not zero. Algorithm Specification (using a Flowchart) Start Read divisor False divisor == 0 True Write divisor “invalid divisor” Read dividend Write dividend / divisor dividend % divisor Write “Thank you” Stop © 2013 Gilbert Ndjatou Page 69 Figure CS3 Using a Compound Statement in an if-else Structure /******************************************************* ****************** Program to read two integer values and to compute the quotient and the remainder in the division of the second value by the first. ******************************************************** *****************/ #include <iostream> using namespace std;

int main() { int divisor, dividend; // to hold the first value // to hold the second value /*---------------------- read in the divisor----------------------*/ cout << “\n\nEnter the divisor please:\t”; cin >> divisor; /*-----read the dividend and compute the quotient and the remainder-*/ if (divisor == 0) // it is invalid cout << endl << divisor << “\tIS INVALID DIVISOR”; else { cout << “\n\nEnter the dividend please:\t”; cin >> dividend; cout << “\nThe quotient in the division of:\t” << dividend << “ by ” << divisor << “\tis:\t”

<< (dividend / divisor); cout << “\nand the remainder is:\t” << (dividend % divisor); } cout << “\n\nThank you for using this program”; return( 0 ); } Exercise CS6* Assuming that all variables are properly defined and initialized, what are the error(s) in each of the following program segments? a. cin >> num; if (num = 5) sum = num + 4; else sum = num + 10; b. cin >> num; if (num > 5 && <= 10) cout << (num + 15); else cout << (num - 3); © 2013 Gilbert Ndjatou Page 70 c. cin >> num; if (num < 15) result1 = 2 * num; result2 = num + 20;

else result1 = 5 * num; result = result1 + result2; cout << “\nThe result is:\t” << result; Exercise CS7* Trace the execution of the following program segment and show its output for each of the following input values: a. input: 4 b. input: 20 Line # Statements 1 cin >> num; 2 if (num <10 ) { 3 num = num + 6; 4 cout << endl << “num =\t” << num; } else 5 cout << endl << “num / 4 =\t” << (num / 4);

6 cout << endl << “result =\t�<< (2 * num); Exercise CS8 Write a C++ code segment to read an integer value into variable num1and another integer value into variable num2 and to do the following: if the value in variable num1 is greater than the value in variable num2 do the following: add 1 to the value in num1 and 5 to the value in num2 and then print the result of the product of the new value in num1 and the new value in num2; otherwise, do the following: subtract 1from the value in num1 and 5 from the value in num2 and then print the result of the product of the new value in num1 and the new value in num2. Exercise CS9 A department store offers a rebate of 5% if the total amount of the purchase is at least 100 and a rebate of 2% otherwise. A sale tax of 8.25 % is also imposed on the price of each purchase (after the rebate).

Write a C++ code segment to read the unit price and the number of items purchased, and to output the amount of the rebate and the price (including taxes) of the purchase. Š 2013 Gilbert Ndjatou Page 71 Counter-Controlled Iteration using the while Structure The solutions of some problems require the CPU to repeat the execution of one or more statements a certain number of times. Example: Suppose that a candy bar costs 70 cents in a candy machine that only accepts dimes. To buy a candy

bar from this machine, a user has to deposit a dime in the machine seven times. The solutions of these types of problems are written using a counter-controlled iteration (or loop) which is specified using a flowchart as follows: counter = initial-value False Counter < final-value True Body-of-the-Loop NextStatement Body-of- the- Loop counter consists of the statements whose executions are repeated by the CPU. is called the loop counter it is a variable that is used to hold the number of repetitions NextStatement is the first statement to be executed after the repetitions.

Initial/Final Value If the initial value is 0 (no execution is done yet), then the final value must be the number of executions. If the initial value is 1 (one execution is about to be done), then the final value must be the number of execution plus 1. Loop Increment After each execution of the body-of-the-loop, the loop counter must be updated to reflect the number of executions.

The statement that you use to update the loop counter is called the loop increment.

The loop increment is a statement in the body-of-the-loop. Š 2013 Gilbert Ndjatou Page 72 Algorithm of the Candy Bar Machine (using a flowchart) Start total = 0 count = 0 False True count < 7 Read coin total = total + coin Write

“Thank you” count = count + 1 Stop Specification of a CounterControlled Iteration in C++ A counter-controlled iteration may be specified in C++ using the while structure as follows: counter = initial-value; while (counter < final-value) { <Body-of-of the-loop (including the increment-statement)> } <Next-statement> © 2013 Gilbert Ndjatou Page 73 Algorithm for the Candy Bar Machine Problem (in C++) int total, coin, count; // to hold the total value of the coins // to hold the value of a coin // to hold the number of repetitions total = 0; count = 0; while ( count < 7 ) { cin >> coin; total += coin; count ++; } cout << endl << “Thank You”; Case Study CS4 Problem Statement

Write a program to read 30 weight measurements in pounds and to convert them into kilograms. Note that 1 Lb = .454 Kgs. Program Logic Input: 30 weight measurements (in Lbs). Output: 30 weight measurements (in Kgs). Variables: count (int) to count the weight measurements: Initial value is 0 - Final value is 30. pound (double) © 2013 Gilbert Ndjatou to hold a weight measurement in Lbs. Page 74 Algorithm Specification (using a flowchart) Start count = 0 False True count < 30 Read pound Write .454 * pound count = count + 1 Write “Thank you” Stop Figure CS4 Counter-Controlled Iteration using the while Structure Line Number 1 /******************************************************* ****** 2 Program to read 30 weight measurements in pounds and convert 3 them into kilograms 4 ******************************************************** *****/

5 #include <iostream.h> 6 #include <iomanip.h> 7 #define MAXCOUNT 30 8 #define COEFICIENT .454 9 10 int main() 11 { 12 int count; // to count weight measurements 13 double pound,

// the current weight measurement in Lbs 14 Š 2013 Gilbert Ndjatou Page 75 15 16 17 18 19 20 21 22 23 24 25 26 27 cout.setf(ios :: fixed); // in order to use formatted cout.setf(ios :: showpoint); // floating-point output cout << setprecision(2); // 2 digits after the decimal point /*read all weight measurements (Lbs) and convert them to Kgs*/ count = 0; // no weight measurement is read so far while (count < MAXCOUNT) // repeat thirty times

{ cout << “\nEnter a weight measurement please:\t”; cin >> pound; cout << “\t = ” << (.454 * pound); count ++; } 28 29 27 28 cout << endl return (0); << “Thank you”; } The body of the loop consists of the statements in line 23 to line 26. The loop-counter is initialized in line 20; and the loop-increment statement is in line 26. Exercise CS10* Assuming that all variables are properly defined, find the mistake(s) in the specification of each of the following while structures: a. while( count < 10 ) { cin >> num; cout << 2 * num;

count = count + 1; } b. while(count < 10) { count = 0; cin >> num; cout << 2 * num; count = count + 1; } c. count = 0; while(count < 10 ) { cin >> num; cout << 2 * num; } Exercise CS11 Write a C++ code segment to read 50 temperature values in Fahrenheit and to convert them to Celsius.

You convert a Fahrenheit temperature to Celsius by using the following formula: Celsius = 5.0 / 9 * (Fahrenheit - 32). Š 2013 Gilbert Ndjatou Page 76 Using a Running Total A running total is a variable that is used to compute the sum of the values processed at ach iteration of a loop: it is first initialized to 0 and each new value processed in the loop is added to the previous total. Case Study CS5 Problem Statement Write a program to read 20 integer values and to compute their sum.

Program Logic Input: 20 integer values. Output: their sum. Variable: count (int) value to count the values: Initial value is 0 (no value is read so far); Final value is 20. (int) totalValue (int) to hold the integer value read. to hold the sum of the integer values read so far: Initial value is 0.

Algorithm Specification (using a Flowchart) Start totalValue = 0 count = 0 False count < 20 True Read value totalValue = totalValue + value Write totalValue count = count + 1 Stop Š 2013 Gilbert Ndjatou Page 77 Figure CS5 Counter-Controlled Iteration using the while Structure Line Number 1 /******************************************************* ****** 2 Program to read twenty integer values and to compute their sum.

3 ******************************************************** *****/ 4 #include <iostream> 5 using namespace std;

6 #define MAXCOUNT 20 7 8 int main() 9 { 10 int count; // to count values 11 value, // to hold the value read 12 totalValue; // the hold the sum of the values read so far 13 14 /*------read all values and compute their sum-------*/ 15 totalValue = 0;

15 count = 0; // no value has been read so far 16 while (count < MAXCOUNT) // repeat twenty times 17 { 18 cout << “\nEnter an integer value please:\t�; 19 cin >> value; 20 totalValue += value; 21 count++; 22 } 23

24 /*-----print the sum of all the values read----------*/ 25 cout << “\n\nThe sum of all the values read is:\t�

26 << totalValue; 27 return (0); 28 } Exercise CS12* Each of the following code segments is supposed to read 10 integer values and to compute their sum.

What is wrong in each of these code..