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Solution Manual for Absolute C++, 6th Edition

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual Chapter 1 C++ Basics Key Terms functions program int main() return 0 identifier case - sensitive keyword or reserved word declare floating - point number fixed width integer types auto unsigned assignment statement uninitia lized variable assigning int values to double variables mixing types integers and Booleans literal constant scientific notation o r floating - point notation quotes C - string string escape sequence const modifier declared constant mixing types precedence rules integer division the % operator nega tive integers in division type cast type coercion increment operator decrement operato r v++ versus ++v cout expression in a cout statement spaces in output newline character deciding between \ n and endl format for double values

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual magic formula outputting money amounts cerr cin how cin works sep arate numbers with spaces when to comment #include , preprocessor namespace using namespace Brief Outline 1.1 Introduction to C++ Origins of the C++ Language C++ and Object - Oriented Programming The Character of C++ C++ Terminology A Sample C++ Program 1.2 Variables, Expressions, and Assignment Statements Identifiers Variables Assignment Statements More Assignment Statements Assignment Compatibility Literals Escape Sequences Naming Constants Introduction to the string class Arithmetic Operators and Expressions Integer and Floating - Point Division Type Casting Increment and Decrement Operators 1.3 Console Input/Output Output Using cout New Lines in Output Formatting for Numbers with a Decimal Point Output with cerr Input Using cin 1.4 Program Style Comments 1.5 Libraries and Namespaces Libraries and include Directives Namespaces

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual 1. Introduction and Teaching Suggestions This chapter introduces the students to the history of the C++ language and begins to tell them about what types of programs can be written in C++ as well as the basic structure of a C++ program. During the discussions on compilation and running a program, care should be taken to explain the process on the particular computer system that the students will be using, as different computing/development environments will each have their own specific directions that will need to be followed. In the development of this instructor’s manual, a majority of the programs have been compiled using g ++ 4. 4.7 on Ubuntu Linux, g++ 3.4 on cygwin, and Visual Studio .NET 2013 . There are significant differences between the development environments and sometimes on the compilers as well. This is especially the case with C++11 where command line options may or may not be needed to compile, and some libraries may be unavailable for later sections (e.g. threading, regular expressions). Simple programming elements are then introduced, starting with simple variable declarations , data types, assignment statements, and eventually evolving into arithmetic expressions . String variable s are not introduced in detail until Chapter 9, but an introduction is given and could be elaborating upon if desired . If time allows, a discussion of how the computer stores data is appropriate. While some of the operations on the primitives are familiar to students, operations like modul u s (%) are usually not and require additional explanation. Also, the functionality of the increment and decrement operators requires attention. The issue of type casting is also introduced, which syntactically as well as conceptually can be difficult for students. Some students that have previously learned C may use the old form of type casting (e.g. (int)), but should be encouraged to use the newer form (e.g. static_cast<int>). The section on programming style further introduces the ideas of conventions for naming of programmatic entities and the use and importance of commenting source code. Commenting is a skill that students will need to develop and they should begin commenting their code from the first program that they complete. Indentation is also discussed. However, many development environments actually handle this automatically. 2. Key Points Compiler . The compiler is the program that translates source code into a language that a computer can understand. Students should be exposed to how compiling works in the ir particular development environment. If using an IDE, it is often instructive to show command - line compiling so students can a sense of a separate program being invoked to translate their code into machine code. This process can seem “magical” when a button is simply pressed in an IDE to compile a program. Syntax and Semantics. When discussing any programming language, we describe both the rules for writing the language, i.e. its grammar, as well as the interpretation of what has been written , i.e. its semantics . For syntax, we have a compiler that will tell us when we have made a mistake. We can correct the error and try compiling again. However, the bigger challenge may

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual lie in the understanding of what the code actually means. There is no “compiler” for telling us if the code that is written will do what we want it to, and this is when the code does not do what we want, it most often takes longer to fix than a simple syntax error. Names (Identifiers). C++ has specific rules for how you can name an entity in a program. These rules are compiler enforced, but students should be able to recognize a correct or incorrect identifier. Also, there are common conventions for how C++ names its programming entities. Variable names begin with a l ower case letter while constants are in all upper case . However, these conventions are not compiler enforced. The book and the source code for C++ itself use these conventions and it is helpful for students to understand that if they follow them, their code is easier for others to read. Variable Declarations. C++ requires that all variables be declared before they are used. The declaration consists of the type of the variable as well as the name. You can declare more than one variable per line. Assignment Statements with Primitive Types. To assign a value to a variable whose type is a primitive, we use the assignment operator, which is the equals (=) sign. Assignment occurs by first evaluating the expression on the right hand side of the equals sign and then assigning the value to the variable on the left. Confusion usually arises for students when assigning the value of one variable to another. Showing that x = y is not the same as y = x is helpful when trying to clear up this confusion. Initializing a Variable in a Declaration . We can and should give our variables an initial value when they are declared. This is achieved through the use of the assignment operator. We can assign each variable a value on separate lines or we can do mul tiple assignments in one line . Assignment Compatibility . Normally, we can only assign values to a variable that are of the same type as we declared the variable to be. For example, we can assign an integer value to an integer variable. However, we can also assign a char value to an integer due to the following ordering: char  short  int  long  float  double Values on the left can be assigned to variables whose types are to the right. You cannot go in the other direction. In fact, the compiler will give an error if you do. However, you may receive a compiler warning message about loss of precision. What is Doubled? This discussion concerns how floating - point numbers are stored inside the computer. A related topic would be to show the conversion of these numbers into the format (e.g. IEEE 754 into two’s complement) that the computer uses. Escape Sequences. When outputting strings, the \ character is used to escape the following character and interpret it literally. It is useful to use this to show how to output " or \ along with untypable characters, such as newlines or tabs.

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual I/O with cin, cout, cerr. This discussion shows how to input values from the keyboard and output them to the screen. Under a Unix system it is easy to show the difference between cout and cerr by redirecting the output. Namespaces. This section illustrates how functions can exist within different namespaces. Students that have previously learned C sometimes have difficulty with namespaces. Some instructors wish to avoid “using namespace std;” and instead prefer to start their programs only using the constructs that are actually used (e.g. using std::cin;). The book starts using the entire namespace but gravitates toward the latter toward the end of the book. Naming Constants. Program style is important and varies from one person to another. However, having programmatic style standards makes programs easier to read for everyone. One of these style points can be the naming of constants in the program. The convention that is introduced is the one that is common to C++ and the text. 3. Tips Error Messages. One of the most frustrating parts of learning how to program is learning to understand the error messages of the compiler. These errors, which are commonly called syntax errors, frustrate students. It is helpful to show some common error messages from the compiler so that students have a frame of reference when they see the errors again themselves. Also important for students to note is that even though C++ states the line number that the error occurred on, it is not always accurate. Run - time errors occur when the program has been run. For this section, creating a simple statement that divides by zero can generate one such error. The last type of error, a logic error is one that is hardest to spot because on the surface the program runs fine, but does not produce the correct result. The difference between x++ and ++x could be used here to illustrate a logic error. 4. Pitfalls Uninitialized Variables. Variables that are declared but not assigned a value are uninitialized. It is good programming practice to always initialize your variables. It may be instructive to output the contents of uninitialized variables to show the unpredictable values they may contain. Uninitialized variables used in computation can cause errors in your program and the best way to avoid these errors is to always give variables an initial value. This can most easily be done right when the variable is declared. Round - off Errors in Floating - Point Numbers. One of the places to show the fallibility of computers is in the round - off errors that we experience when using floating point numbers. This topic relates to why the type is named double and also deals with the representation of floating point numbers in the system. This problem occurs because not all floating - point numbers are finite, a common example being the decimal representation of the fraction 1/3. Because we can only store so many digits after the deci mal points, our computation is not always accurate. A

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual discussion of when this type of round off error could be a problem would be appropriate to highlight some of the shortcomings of computing. Division with Whole Numbers . In C++ , all of the arithmetic operations are closed within their types. Therefore, division of two integers will produce an integer, which will produce an answer that most students do not expect. For example, the integer 1 divided by the integer 2 will produce the integer 0. Students will expect 0.5. One way to get a floating - point answer out of integer division is to use typecasting. Another way is to force floating - point division by making one of the integers a floating - point number by placing a “.0” at the end. Experimentation with this issue is important to show the different results that can be obtained from integer division. Order of Evaluation . The order of evaluation of subexpressions is not guaranteed. For example in (n + (++n)) there will be a different result if ++n is evaluated first or second. The best advice is to avoid such scenarios. Precedence of operators is discussed in chapter 2. 5. Programming Projects Answers 1. Metric - English units Conversion A metric ton is 35,273.92 ounces. Write a C++ program to read the weight of a box of cereal in ounces then output this weight in metric tons, along with the number of boxes to yield a metric ton of cereal. Design: To convert 14 ounces (of cereal) to metric tons, we use the 'ratio of units' to tell us whether to divide or multiply: 1 metric tons 14 ounces * = 0.000397 metric tons 35,273.92 ounce The explicit use of units will simplify the determination of whether to divide or to multiply in making a conversion. Notice that ounces/ounce becomes unit - less, so that we are left with metric ton units. The number of ounces will be very, very much larger than the number of metric tons. It is then reasonable to divide the number of ounces by the number of ounces in a metric ton to get the number of metric tons. Let metricTonsPerBox be the weight of the cereal contained in the box in metric tons, and let ouncesPerBox be the weight of the cereal contained in the box in ounces. Then in C++ the formula becomes: const double ouncesPerMetricTon = 35272.92; metricTonsPerBox = ouncesPerBox/ouncesPerMetricTon; This is metric tons PER BOX, whence the number of BOX(es) PER metric ton should be the reciprocal of this number:

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual boxesPerMetricTon = 1 / metricTonsPerBox; Once this analysis is made, the code proceeds quickly: // Purpose: To convert cereal box weight from ounces to // metric tons and to compute number of boxes of cereal that // constitute a metric ton of cereal. #include <iostream> using namespace std; const double oncesPerMetricTon = 35272.92; int main() { double ouncesPerBox, metricTonsPerBox, boxesPerMetricTon; cout << “enter the weight in ounces of your” << “favorite cereal:” <<endl; cin >> ouncesPerBox; metricTonsPerBox = ouncesPerBox / oncesPerMetricTon; boxesPerMetricTon = 1 / metricTonsPerBox; cout << "metric tons per box = " << metricTonsPerBox << endl; cout << "boxes to yield a metric ton = " << boxesPerMetricTon << endl; return 0; } A sample run follows: ~/AW$ a.out enter the weight in ounces of your favorite cereal: 14 metric tons per box = 0.000396905 boxes to yield a metric ton = 2519.49 enter the weight in ounces of your favorite cereal: 2. Lethal Dose Certain artificial sweeteners are poisonous at some dosage level. It is desired to know how much soda a dieter can drink without dying. The problem statement gives no information about how to scale the amount of toxicity from the dimensions of the experimental mouse to the dimensions of the dieter. Hence the student must supply some more or less reasonable assumption as basis for the calculation.

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual This solution supposes the lethal dose is directly proportional to the weight of the subject, hence lethal_dose_dieter = lethal_dose_mouse * weight of dieter weight of mouse _ _ _ _ This program accepts weight of a lethal dose of sweetener for a mouse, the weight of the mouse, and the weight of the dieter, then calculates the amount of sweetener that will just kill the dieter, based on the lethal dose for a mouse in the lab. If the student has problems with grams and pounds, a pound is 454 grams. It is interesting that the result probably wanted is a safe number of cans, while all the data can provide is the minimum lethal number. Some students will probably realize this, but my experience is that most will not, or will not care. I just weighed a can of diet pop and subtracted the weight of an empty can. The result is about 350 grams. The label claims 355 ml, which weighs very nearly 355 grams. To get the lethal number of cans from the number of grams of sweetener, you need the number of grams of sweetener in a can of pop, and the concentration of sweetener, which the problem assumes 0.1% , that is, a conversion factor of 0.001. gramsSweetenerPerCan = 350 * 0.001 = 0.35 grams/can cans = lethalDoseForDieter / (0.35 grams / can) //Input: lethal dose of sweetener for a lab mouse, weights // of mouse and dieter, and concentration of sweetener in a // soda. //Output: lethal dose of soda as a number of cans. //Assumption: lethal dose proportional to weight of subject // concentration of sweetener in the soda is 1/10 percent #include <iostream> using namespace std; const double concentration = .001; // 1/10 of 1 percent const double canWeight = 350; const double gramsSweetenerPerCan = canWeight * concentration; //units: grams/can int main() { double lethalDoseMouse, lethalDoseDieter, weightMouse, weightDieter; //units: grams double cans; cout << "Enter the weight of the mouse in grams" << endl; cin >> weightMouse; cout << "Enter the lethal dose for the mouse in“ << ” grams " << endl; cin >> lethalDoseMouse; cout << "Enter the desired weight of the dieter in”

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual <<“ grams " << endl; cin >> weightDieter; lethalDoseDieter = lethalDoseMouse  weightDieter/weightMouse; cout << "For these parameters: \ nmouse weight: " << weightMouse << " grams " << endl << "lethal dose for the mouse: " << lethalDoseMouse << " grams" << endl << "Dieter weight: " << weightDieter << " grams " << endl << "The lethal dose in grams of sweetener is: " << lethalDoseDieter << endl; cans = lethalDoseDieter / gramsSweetenerPerCan; cout << "Lethal number of cans of pop: " << cans << endl; return 0; } A typical run follows: Enter the weight of the mouse in grams 15 Enter the lethal dose for the mouse in grams 100 Enter the desired weight of the dieter, in grams 45400 For these parameters: mouse weight: 15 grams lethal dose for the mouse: 100 grams Dieter weight: 45400 grams The lethal dose in grams of sweetener is: 302667 Lethal number of cans of pop: 864762 3. Pay Increase The workers have won a 7.6% pay increase, effective 6 months retroactively. This program is to accept the previous salary, then outputs the retroactive pay due the employee, the new annual salary, and the new monthly salary. Allow user to repeat as desired. The appropriate formulae are: newSalary = salary * ( 1 + INCREASE ); monthly = salary / 12; retroactive = (newSalary – salary) / 2;

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual The code follows: //Given 6 months retroactive, 7.6% pay increase, //Input: old salary //Output: new annual and monthly salary, retroactive pay due #include <iostream> using namespace std; const double INCREASE = 0.076; int main() { double oldSalary, salary, monthly, retroactive; char ans; cout << "Enter current annual salary." << endl << "I'll return new annual salary, monthly ” << “salary, and retroactive pay." << endl; cin >> oldSalary; salary = oldSalary * ( 1 + INCREASE ); monthly = salary / 12; retroactive = (salary – oldSalary) / 2; cout << "new annual salary " << salary << endl; cout << "new monthly salary " << monthly << endl; cout << "retroactive salary due: " << retroactive << endl; return 0; } 17:50:12:~/AW$ a.out Enter current annual salary. 100000 I'll return new annual salary, monthly salary, and retroactive pay. new annual salary 107600 new monthly salary 8966.67 retroactive salary due: 53800 4. Consumer Loan This problem is an example where the student needs to have a solution in hand before going to the computer. Algebra will solve the problem based on what is given. First compute the “discounted loan face value” in terms of the other things known: InterestRate and NoYearsToRun : AmountRecevied = FaceValue – discount Discount = FaceValue * AnnualInterestRate * NoYearsToRun; AmountReceived = FaceValue – FaceValue*AnnualInterestRate*NoYearsToRun Then ask them to solve for FaceValue :

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual Fac eValue = AmountReceived / (1 AnnualInterestRate*NoYearsToRun) Writing a program for this will be easy. 5. Occupancy of Meeting Room beyond file - law limits. Program reads the maximum occupancy, then the then the number of attendees. If the maximum has been exceeded, announces either that the room has its maximum occupancy exceeded and how many must leave, else it announces that the maximum has not been exceeded. This problem uses the if - else illustrated in Display 1.1, and a one - sided if statement, //Maximum Occupancy #include <iostream> using namespace std; int main() { int maxOccupancy; int numberOccupants; cout << "Enter the Maximum occupancy for the room. \ n"; cin >> maxOccupancy; cout << maxOccupancy << endl; cout << "Enter the number of occupants of the room. \ n"; cin >> numberOccupants; cout << numberOccupants << endl; if(numberOccupants > maxOccupancy) cout << "ATTENTION: MAXIMUM OCCUPANCY EXCEEDED. \ n" << "THE LAW REQUIRES " << numberOccupants - maxOccupancy << " PERSONS TO LEAVE THE ROOM IMMEDIATELY \ n"; else cout <<"The number of occupants does not exceed " << "the legal maximum. \ n"; return 0; } /* A typical run is: Enter the Maximum occupancy for the room. 250 Enter the number of occupants of the room. 375 ATTENTION: MAXIMUM OCCUPANCY EXCEEDED. THE LAW REQUIRES 125 PERSONS TO LEAVE THE ROOM IMMEDIATELY Another run: Enter the Maximum occupancy for the room. 250 Enter the number of occupants of the room. 225 The number of occupants does not exceed the legal maximum. */

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual 6. Overtime Pay This problem uses the if - else illustrated in Display 1.1, and a one - sided if statement, which is just an if - else with the else clause omitted. //pay roll problem: //Inputs: hoursWorked, number of dependents //Outputs: gross pay, each deduction, net pay // //Outline: // //regular payRate = $16.78/hour for hoursWorked <= 40 //hours. //For hoursWorked > 40 hours, // overtimePay = (hoursWorked - 40) * 1.5 * payRate. //FICA (social security) tax rate is 6% //Federal income tax rate is 14%. //Union dues = $10/week . //If number of dependents >= 3 // $35 more is withheld for dependent health insurance. // #include <iostream> using namespace std; const double PAYRATE = 16.78; const double SS_TAX_RATE = 0.06; const double FedIRS_RATE = 0.14; const double STATETAX_RATE = 0.05; const double UNION_DUES = 10.0; const double OVERTIME_FACTOR = 1.5; const double HEALTH_INSURANCE = 35.0; int main() { double hoursWorked, grossPay, overTime, fica, incomeTax, stateTax, netPay; int numberDependents, employeeNumber; //set the output to two places, and force .00 for cents cout.setf(ios::showpoint); cout.setf(ios::fixed); cout.precision(2); // compute payroll cout << "Enter employee SSN (digits only," << " no spaces or dashes) \ n"; cin >> employeeNumber ; cout << endl << employeeNumber << endl; cout << "Please enter hours worked \ n"; cin >> hoursWorked; cout << endl << hoursWorked << endl; cout << "Please enter number of dependants." << endl; cin >> numberDependents; cout << endl << numberDependents << endl << endl; if (hoursWorked <= 40 )

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual grossPay = hoursWorked * PAYRATE; else overTime = (hoursWorked - 40) * PAYRATE * OVERTIME_FACTOR; if (hoursWorked > 40) grossPay = 40 * PAYRATE + overTime; fica = grossPay * SS_TAX_RATE; incomeTax = grossPay * FedIRS_RATE; stateTax = grossPay * STATETAX_RATE; netPay = grossPay - fica - incomeTax - UNION_DUES - stateTax; if (numberDependents >= 3) netPay = netPay - HEALTH_INSURANCE; //now print report for this employee: cout << "Employee number: " << employeeNumber << endl; cout << "hours worked: " << hoursWorked << endl; cout << "regular pay rate: " << PAYRATE << endl; if (hoursWorked > 40) cout << "overtime hours worked: " << hoursWorked - 40 << endl; if (hoursWorked > 40) cout << "with overtime premium: " << OVERTIME_FACTOR << endl; cout << "gross pay: " << grossPay << endl; cout << "FICA tax withheld: " << fica << endl; cout << "Federal Income Tax withheld: " << incomeTax << endl; cout << "State Tax withheld: " << stateTax << endl; if (numberDependents >= 3) cout << "Health Insurance Premium withheld: " << HEALTH_INSURANCE << endl; cout << "Flabbergaster's Union Dues withheld: " << UNION_DUES << endl; cout << "Net Pay: " << netPay << endl << endl; return 0; } /* Two typical runs follow: Enter employee SSN (digits only, no spaces or dashes) 234567890 234567890 Please enter hours worked 37.00 37.00 Please enter number of dependants.

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual 1 1 Employee number: 234567890 hours worked: 37.00 regular pay rate: 16.78 gross pay: 620.86 FICA tax withheld: 37.25 Federal Income Tax withheld: 86.92 State Tax withheld: 31.04 Flabbergaster's Union Dues withheld: 10.00 Net Pay: 455.64 Enter employee SSN (digits only, no spaces or dashes) 234567890 234567890 Please enter hours worked 54.00 54.00 Please enter number of dependants. 4 4 Employee number: 234567890 hours worked: 54.00 regular pay rate: 16.78 overtime hours worked: 14.00 with overtime premium: 1.50 gross pay: 1023.58 FICA tax withheld: 61.41 Federal Income Tax withheld: 143.30 State Tax withheld: 51.18 Health Insurance Premium withheld: 35.00 Flabbergaster's Union Dues withheld: 10.00 Net Pay: 722.69 /* 7. Calories One way to measure the amount of energy that is expended during exercise is to use metabolic equivalents (MET). Here are some METS for various activities: Running 6 MPH: 10 METS Basketball: 8 METS Sleeping: 1 MET The number of calories burned per minute may be estimated using the formula: Calories/Minute = 0.0175  MET  Weight(Kg)

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual Write a program that inputs a subject’s weight in pounds, the number of METS for an activity, and the num ber of minutes spent in that activity, and then outputs the estimate for total number of calories burned. One kilogram is equal to 2.2 pounds. //calories.cpp //This program calculates the amount of energy expended // using the concept of metabolic equivalents. The formula // used is Calories/Minute = 0.0175 * MET * WeightInKilos //One Kg = 2.2 Pounds #include <iostream> #include <cstdlib> using namespace std; const double POUNDS_TO_KG = 1 / 2.2; // ====================== // main function // ====================== int main() { // // Variable declarations double mets; double weight_lb; double mins; double weight_kg; double calories; cout << endl << "Welcome to the calorie calculator." << endl; // -------------------------------- // ----- ENTER YOUR CODE HERE ----- // -------------------------------- cout << "Enter your weight in pounds: " << endl; cin >> weight_lb; cout << "Enter the number of METS for the activity: " << endl; cin >> mets; cout << "Enter the number of minutes spent exercising: " << endl; cin >> mins; // Convert from pounds to kilograms weight_kg = POUNDS_TO_KG * weight_lb; // Formula to compute calories calories = 0.0175 * mets * weight_kg * mins; // Output total calories cout << "You burned an estimated " << calories << " calories." << endl << endl;

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual // -------------------------------- // --------- END USER CODE -------- // -------------------------------- 8. Babylonian The Babylonian algorithm to compute the square root of a number n is as follows: 1. Make a guess at the answer (you can pick n/2 as your initial guess). 2. Compute r = n / guess 3. Set guess = (guess +r) / 2 4. Go back to step 2 for as many iterations as necessary. The more that steps 2 and 3 are repeated, the closer guess will become to the square root of n . Write a program that inputs an integer for n , iterates through the Babylonian algorithm five times, and outputs the answer as a double to two decimal places. Your answer will be most accurate for small values of n . CodeMate Hints: Make guess a double CodeMate Hints: Typecast using static_cast<double>(n) described on page 23 CodeMate Hints: See page 31 for the magic formula on setting the precision for outputting doubles. //babylonian.cpp // //This program uses the Babylonian algorithm, using five // iterations, to estimate the square root of a number n. #include <iostream> #include <cstdlib> using namespace std; // ====================== // main function // ====================== int main() { // -------------------------------- // ----- ENTER YOUR CODE HERE ----- // -------------------------------- // Variable declarations double guess; int n; double r; cout << endl << "This program makes a rough estimate for square roots." << endl; cout << "Enter an integer to estimate the square root of: " << endl; cin >> n;

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual // Initial guess guess = static_cast<double>(n)/2; // First guess r = static_cast<double>(n)/ guess; guess = (guess+r)/2; // Second guess r = static_cast<double>(n)/ guess; guess = (guess+r)/2; // Third guess r = static_cast<double>(n)/ guess; guess = (guess+r)/2; // Fourth guess r = static_cast<double>(n)/ guess; guess = (guess+r)/2; // Fifth guess r = static_cast<double>(n)/ guess; guess = (guess+r)/2; // Code to set the precision to 2 decimal places cout.setf(ios::fixed); cout.setf(ios::showpoint); cout.precision(2); // Output the fifth guess cout << "The estimated square root of " << n << " is " << guess << endl << endl; // -------------------------------- // --------- END USER CODE -------- // -------------------------------- 9. Coupons The video game machines at your local arcade output coupons depending upon how well you play the game. You can redeem 10 coupons for a candy bar or 3 coupons for a gumball. You prefer candy bars to gumballs. Write a program that inputs the number of coupons you win and outputs how many candy bars and gumballs you can get if you spend all of your coupons on candy bars first, and any remaining coupons on gumballs. CodeMate Hints: Use the % operator to compute the number of coupons remaining //coupons.cpp // // This program computes the number of candy bars and gumballs you // can get by redeeming coupons at an arcade. 10 coupons is // redeemable for candy bars and 3 coupons for gumballs. You // would like as many candy bars as possible and only use // remaining coupons on gumballs.

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual #include <iostream> #include <cstdlib> using namespace std; // ====================== // main function // ====================== int main() { // Variable declarations int num_coupons = 0; int num_coupons_after_candybars = 0; int num_coupons_after_gumballs = 0; int num_candy_bars = 0; int num_gumballs = 0; cout << endl << "Candy calculator. Enter number of coupons to redeem:" << endl; // // -------------------------------- // ----- ENTER YOUR CODE HERE ----- // -------------------------------- cin >> num_coupons; // Integer division below discards any remainder num_candy_bars = num_coupons / 10; // Calculate remaining coupons num_coupons_after_candybars = num_coupons % 10; // Calculate gumballs num_gumballs = num_coupons_after_candybars / 3; // Calculate any leftover coupons num_coupons_after_gumballs = num_coupons_after_candybars % 3; // Output the number of candy bars and gumballs cout << "Your " << num_coupons << " coupons can be redeemed for " << num_candy_bars << " candy bars and " << num_gumballs << " gumballs with " << num_coupons_after_gumballs << " coupons leftover." << endl << endl; // -------------------------------- // --------- END USER CODE -------- // -------------------------------- 10. Freefall

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual Write a program that allows the user to enter a time in seconds and then outputs how far an object would drop if it is in freefall for that length of time. Assume no friction or resistance from air and a constant acceleration of 32 feet per second due to gravity. Use the equation: 2 2 1 distance time on accelerati    // Programming Project 1.10 #include <iostream> using namespace std; int main() { cout << "Enter a time in seconds." << endl; int s; cin >> s; int distance; distance = (32 / 2) * (s * s); cout << "An object in freefall for " << s << " seconds will fall " << distance << " feet." << endl; // Type a key and enter to close the program char c; cin >> c; } 11. Time Conversion Write a program that inputs an integer that represents a length of time in seconds. The program should then output the number of hours, minutes, and seconds that corresponds to that number of seconds. For example, if the user inputs 50391 total seconds then t he program should output 13 hours, 59 minutes, and 51 seconds. // Programming Project 1.11 #include <iostream> using namespace std; int main() { cout << "Enter a time in seconds." << endl; int s; cin >> s; int hours, minutes, seconds; hours = s / 3600; minutes = (s % 3600) / 60; seconds = (s % 3600) % 60; cout << s << " total seconds is equivalent to " << hours << " hours, " << minutes <<

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual " minutes, and " << seconds << " seconds. "<< endl; // Type a key and enter to close the program char c; cin >> c; } 12. Ideal Body Weight Estimation A simple rule to estimate your ideal body weight is to allow 110 pounds for the first 5 feet of height and 5 pounds for each additional inch. Write a program with a variable for the height of a person in feet and another variable for the additional inches and input values for these variables from the keyboard. Assume the person is at least 5 feet tall. For example, a person that is 6 feet and 3 inches tall would be represented with a variable that stores the number 6 and another variable that stores the number 3. Based on these values calculate and output the ideal body weight. / / Programming Project 1.12 #include <iostream> using namespace std; int main() { int heightFeet = 5; int heightInches = 5; cout << "Your ideal body weight is "; int weight = 110 + ((heightFeet - 5)*12 + heightInches) * 5; cout << weight << " pounds." << endl; return 0; } 1 3 . Lethal Caffeine Scientists estimate that consuming roughly 10 grams of caffeine at once is a lethal overdose. Write a program that inputs the number of milligrams of caffeine in a drink and outputs how many drinks it would take to kill a person. A 12 ounce can of cola has approximately 34 mg of caffeine while a 16 ounce cup of coffee has approximately 160 mg of caffeine. // Programming Project 1.13 #include <iostream> using namespace std; int main() { int mg; cout << "How many milligrams of caffeine in the drink?" << endl; cin >> mg; // 10 grams to kill a person is 10000 mg

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    Savitch, Absolute C++ 6 /e : Chapter 1, Instructor’s Manual int numDrinks = 10000 / mg; cout << "The number of drinks it will take to kill someone is " << numDrinks << " drinks." << endl; return 0; }

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual Chapter 2 Flow of Control Key Terms Boolean expression && means “and” || means “or” truth tables parentheses precedence rules higher precedence short - circuit complete evaluation integers convert to bool if - else statement parentheses if - else with multiple statements compound statement if statement indenting multiway if - else switch statement controlling expression break state ment default enumeration type conditional operator condit ional operator expression loop body iteration while and do - while compared exe cuting the body zero times comma operator comma expression for statement empty statement infinite loop continue statement ifstream stream file input text file

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual Brief Outline 2.1 Boolean Expressions Building Boolean Expressions Evaluating Boolean Expressions Precedence Rules 2.2 Branching Mechanisms If - else Statements Compound Statements Omitting the else Nested Statements Multiway if - else Statement The switch Statement Enumeration Types The Conditional Operator 2.3 Loops The while and do - while statements Increment and Decrement Operators Revisited The Comma Operator The for statement The break and continue statements Nested Loops 2.4 Introduction to File Input Reading from a Text File Using ifstream 1. Introduction and Teaching Suggestions Without flow of control, a language is not able to make decisions. The ability to choose a path through code based on data is what gives a computer its power. Two flow of control issues not covered in this chapter are function calls and exception handling. Function calls are covered in Chapter 3 and exceptions in Chapter 18. This chapter discusses flow of control using both selection and iteration. The if - statement and loops are introduced both in this chapter. With both topics in one chapter, it is conceivable that this chapter will take longer for students to work through than many of the others in the text. Branching, or selection, is introduced using the if - else statement, the if statement, and the switch statement. Multiple paths and therefore, nested if - else statements are introduced. As the last form of selection available, the conditional operator is introduced at the end of the chapter. Although many students choose not to use this operator, it is useful to cover so students can read code that does use it. With the introduction of looping and selection, students can write fairly complex programs at the end of this chapter. However, what usually begins to happen is that students will write code that has errors. Finding these errors can be tedious and time - consuming. The sections on common pitfalls (infinite loops, semicolon at the end of for loops, etc.) should be covered as an

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual introduction to debugging. An option that would be useful to them if it is available would be to introduce and discuss the use of your computing environment’s debugger at this time. The section on file input allows students to write programs that use real - world data. This is significant because it allows students to move beyond toy problems. The book has several problems based on a list of English words. Although this section requires some “magic” code that is not fully explained until later, students should be able to grasp the basic concepts of reading from a text file and can begin writing program that o perate on large amounts of data. 2. Key Points Boolean Expressions and Relational Operators . The Boolean operators && and || are introduced along with ! and their use with ==, !=, <, >, <=, and >=. Truth tables and building complex Boolean expressions are important for the students to learn how to construct. In addition, students must also learn the precedence rules associated with these operators. If - else statement & Multiway if - else Statement . There are many ways to give the program a sense of choice or branching. First, we can use an if - statement by itself without an else. This allows us the option to do something or skip over it. We can also use an if - else statement, which allows us to take one path or another. Lastly, we can use combinations of these to have more than two choices for execution. As the number of paths increase, so does the complexity of code for the students. Students should be able to follow as well as write these more complicated branching code segments. The switch Statement. The switch also allows for branching, but it has limitations as to what the condition for branching can be. Also, the syntax contains more keywords and is more structured than the if - else. Discussion of the break statement is needed here as the switch will not function properly without the correct use of break between the cases. true and false are n umbers. T rue and false can be represented as the numbers 1 and 0. This is sometimes used by setting a variable equal to the result of an Boolean expression, or by testing a variable inside an if - statement. Syntax for while and do - while Statements. The while and do - while loops are the indefinite loops supported by C++ . They also illustrate the differences between an entry - test and an exit - test loop. The for Statement. The for loop is a definite loop or counting loop that is also an entry - test loop. The syntax for the for loop is different from the other two loops and has a loop counter built right into the construct. However, in C++ , we can have more than one statement inside the parts of the for - loop separated by commas and we can also leave parts empty, which can create many different results when using a for - loop.

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual break and continue. The difference between break and continue are sometimes confused by students, particularly scenarios where continue is useful. It can be instructive to show how the same effect can be achieved by code using if - statements instead of break and continue. Enumeration Types . These types allow us to define a sequence of constant values that are often used in loops or switch statements. They map to integers. 3. Tips Use switch statements for menus . The switch statement’s advantage over the if - statement is increased clarity, especially for large numbers of items that must be compared. This makes it a good choice for implementing menus, in particular when used with functions introduced in Chapter 3. Repeat - N Times Loops. A for loop is the best way to implement a count - based loop that repeats N times. 4. Pitfalls Using = in Place of ==. The instructor should note that the very best of programmers fall into this error. The assignment x = 1; is an expression that has a value. The value of this expression is the value transferred to the variable on the left - hand side of the assignment, in this case, 1.. This value could be used by an if or while as a condition, just as any other expression can, for example: z = (x = 1); This, in fact, is a typical C/C++ idiom. While this is used, it can be confusing. Many instructors discourage its use, although some programmers do use it. Some compilers warn if an assignment expression is used as the Boolean expression in an if statement or within a loop. There is a way to get some further help at the cost of a little discipline. If the programmer will consistently write any constant that may appear in a comparison on the left of the == , as in if(12 == x) ...; instead of if(x == 12) ...; then the compiler will catch this pitfall of using = instead of ==. if(12 = x) // error: invalid l - value in assignment ...; Otherwise this is very hard to see, since this looks right. There are also difficult - to - see pitfalls in using only one & instead of && , or one | , instead of || . Each of these is legal C++, so the student won’t be warned, and the program will run, and produce incorrect results. Warn them.

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual Extra Semicolon. If we define an enum type, and omit the required semicolon after the ending curly brace, the error message on some compilers is incomprehensible. On other compilers, it is as simple as “missing semicolon in enum definition.” The reason the enum definition requires a semicolon is that one can define a variable of enum type between the close curly and the semicolon. enum CompassDir {NORTH, SOUTH, EAST, WEST} direction; Then direction has CompassDir type. If the semicolon is omitted, the compiler thinks that anything following the semicolon wants to be an identifier of enum type, and identifies this as an error, and issues an error message. Another Semicolon Pitfall. This error occurs when a student adds a semicolon to the end of an if statement or for loop, e.g.: if(x == 12); x = 0; After this code segment, the variable x has the value 0. The hard - to - see error is the semicolon at the end of the line with the if . The if statement expects to see a single statement after the closing parenthesis. The semicolon produces a null statement between the close parenthesis and the semicolon. This makes the next statement ( x = 0; ) out of the scope of control of the if. If the if has an else , as in if(x == 12); x = 0; else x = 1; Then the error message appears at the else . The syntax error really is the presence of the else , rather than at the extraneous semicolon, the real perpetrator. As above, the semicolon produces a null statement between the close parenthesis and the semicolon. This makes the next statement ( x = 0; ) out of the scope of control of the if , and makes the else into an “ else without an if” error. The text has a Pitfall section on the extra semicolon in a for loop and another that notes that the extra semicolon problem can cause an infinite loop, for example, x = 10; while(x > 0); { cout << x << endl; x -- ; } A program that has this loop in it will hang. As before, the extraneous semicolon after the while causes the problem. Note that placing assignments in the control of an if or a while is C (and C++ ) idiom. Even so, some compilers warn about this. To stop such runaway programs may require killing the process or hitting control - C. 5. Programming Projects Answers 1. Inflation

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual Write a program to estimate the future cost of an item based on the current inflation rate and the current cost of the item. The inflation rate is to be entered as a percentage, e.g., 5.6 for 5.6 percent. Use a loop to compute the estimated cost based on the (compounded) inflation rate. //inflation estimator //input: item cost in dollars, inflation as a percentage: // for example, 5.6 for 5.6 percent, and an integral number of // years. //output: Estimate in dollars of the cost #include <iostream> using namespace std; int main() { double cost; double rate; // inflation rate int years; //years to time of requested estimate cout << "This is the Future Cost Estimator. \ n" << "Each entry is to be followed by a <return> \ n \ n"; cout << "Enter the cost of the item in dollars: \ n"; cin >> cost; cout << "Enter the inflation rate as a percentage: \ n" << "such as 5.6 for 5.6 percent. "; cin >> rate; rate = rate / 100.0; cout << "Enter an integral number of years to the time \ n" << "when you want the estimate."; cin >> years; for( int i = 0; i < years; i++) cost = (1 + rate) * cost; cout << "The inflated cost is $" << cost << endl; return 0; } 2. Installment Loan Time No down payment, 18 percent / year, payment of $50/month, payment goes first to interest, balance to principal. Write a program that determines the number of months it will take to pay off a $1000 stereo. The following code also outputs the monthly status of the loan. #include <iostream> using namespace std; const double PAYMENT=50.00; int main() { double principal = 1000.; double interest, rate = 0.015; int months = 0;

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual cout << "months \ tinterest \ tprincipal" << endl; cout.setf(ios::fixed); cout.setf(ios::showpoint); cout.precision(2); while(principal > 0) { months++; interest = principal * rate; principal = principal - (PAYMENT - interest); if ( principal > 0 ) cout << months << " \ t \ t" << interest << " \ t \ t" << principal << endl; } cout << "number of payments = " << months; //undo the action that drove principal negative: principal = principal + (PAYMENT - interest); //include interest for last month: interest = principal * 0.015; principal = principal + interest; cout << " last months interest = " << interest; cout << " last payment = " << principal << endl; return 0; } A typical run is: months interest principal 1 15.00 965.00 2 14.47 929.48 3 13.94 893.42 4 13.40 856.82 . . . 21 2.86 143.53 22 2.15 95.68 23 1.44 47.12 number of payments = 24 last months interest = 0.71 last payment = 47.83 3. Chocolate Suppose we can buy a chocolate bar from the vending machine for $1 each. Inside every chocolate bar is a coupon. We can redeem 7 coupons for one chocolate bar from the machine. We would like to know how many chocolate bars can be eaten, including those redeemed via coupon, if we have n dollars. For example, if we have 20 dollars then we can initially buy 20 chocolate bars. This gives us 20 coupons. We can redeem 14 coupons for 2 additional chocolate bars. These two additional chocolate bars have 2 more coupons, so we now have a total of 8 coupons when added to the six leftover from the original purchase. This gives us enough to redeem for one final chocolate bar. As a result we now have 23 chocolate bars and 2 leftover coupons.

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual Write a program that inputs the number of dollars and outputs how many chocolate bars you can collect after spending all your money and redeeming as many coupons as possible. Also output the number of leftover coupons. The easiest way to solve this problem is to use a loop. CodeMate Hint: Use a while loop that redeems coupons for bars and calculates the new number of coupons. Continue the loop as long as you have enough coupons to redeem for a chocolate bar. //chocolate.cpp // //This program computes the number of candy bars we can get. // Suppose we can buy a chocolate bar from the vending machine for $1 // each. Inside every chocolate bar is a coupon. We can redeem 7 // coupons for one chocolate bar from the machine. We would like to // know how many chocolate bars can be eaten, including those redeemed // via coupon, if we have n dollars. // //This program creates a loop that simulates the process by // calculating the number of coupons we have leftover after redeeming // them for candy bars. When we don't have enough coupons to redeem, // the loop stops. #include <iostream> #include <cstdlib> using namespace std; // ====================== // main function // ====================== int main() { // Variable declarations int dollars = 0; int bars = 0; int newbars = 0; int coupons = 0; cout << endl << "Enter the number of dollars you have to spend:" << endl; cin >> dollars; // // -------------------------------- // ----- ENTER YOUR CODE HERE ----- // -------------------------------- // Buy initial candy bars and number of coupons bars = dollars; coupons = bars; // Continually redeem if we have enough coupons while (coupons >= 7)

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    Savitch, Absolute C++ 6/e : Chapter 2 , Instructor’s Manual { // Redeem 7 coupons for one bar newbars = coupons / 7; bars = bars + newbars; // New number of coupons is the amount leftover + // number of coupons we get from the bars we just bought coupons = (coupons % 7) + newbars; } // Output the number of candy bars you can get cout << "You can get " << bars << " candy bars " << " with " << coupons << " coupons leftover." << endl << endl; // -------------------------------- // --------- END USER CODE -------- // -------------------------------- 4. Primes Write a program that finds and prints all of the prime numbers between 3 and 100. A prime number is a number such that one and itself are the only numbers that evenly divide it (e.g., 3,5,7,11,13,17, …). One way to solve this problem is to use a doubly - nested loop. The outer loop can iterate from 3 to 100 while the inner loop checks to see if the counter value for the outer loop is prime. One way to see if number n is prime is to loop from 2 to n - 1 and if any of these numbers evenly divides n then n cannot be prime. If none of the values from 2 to n - 1 evenly divide n , then n must be prime. (Note that there are several easy ways to make this algorithm more efficient). //primes.cpp // //This program calculates the prime numbers between 3 and 100. // It uses a doubly - nested loop. The outer loop iterates from 3 to 100 // while the inner loop checks to see if the counter value for the // outer loop is prime. To check for n to be prime, we loop from 2 to // n - 1 and if any of these numbers evenly divides n then n cannot be // prime. If none of the values from 2 to n - 1 evenly divide n, then n // must be prime. Note that there are several ways to make this // algorithm more efficient. #include <iostream> #include <cstdlib> using namespace std; // ====================== // main function // ====================== int main() { // Variable declarations

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