Chapter 4, Pause Reflect

• 4.1--4.10

• 4.11--4.16

• 4.17--4.23

• 4.24--4.26

• 4.1: Here's a modified version of change.cpp, Program 4.1, that uses the modulus operator %. The statement amount %= 25; can also be written as amount = amount % 25; #include <iostream> using namespace std; // make change in U.S. coins // Owen Astrachan, 1/2/96 int main() { int amount; int quarters, dimes, nickels, pennies; // input phase of program cout << "make change in coins for what amount: "; cin >> amount; // calculate number of quarters, dimes, nickels, pennies quarters = amount/25; amount %= 25; dimes = amount/10; amount %= 10; nickels = amount/5; amount %= 5; pennies = amount; // output phase of program cout << "# quarters =\t" << quarters << endl; cout << "# dimes =\t" << dimes << endl; cout << "# nickels =\t" << nickels << endl; cout << "# pennies =\t" << pennies << endl; return 0; }

• 4.2: If any string other than ``yes'' is entered, the else clause is executed (and the string ``De gustibus non Disputandum'' is printed). The comparisons are case-sensitive so that the string "Yes" is different from "yes".

• 4.3: The variable days is given a default value of 31 so that if the user enters any month with 31 days in it the right answer will be printed. There are five months with 30 days, one with 28 or 29 (February), and six with 31 days. If a default value of 30 days was used, the if/else statements would use if ("january" == month) and so on.

• 4.4: Without knowing the year, it's difficult to determine how many days there are in February (taking leap years into account). The user could be prompted for a year (but the leap year code comes later in the book). One idea is to print that "february has 28 days unless it is a leap year"
• 4.5 and 4.6 (answers are combined) if ("yes" == response) { cout << "Green vegetables are good for you" << endl; cout << "Broccoli is good in Chinese food as well" << endl; } else if ("no" == respone) { cout << "De gustibus non disputandum" << endl; cout << "(There is no accounting for taste)" << endl; } else { cout << "sorry, only responses of yes and no are recognized" << endl; }

• 4.6: see code fragment above

• 4.7: One code fragment that could be used is shown below. There are lots of others. int grade; cout << "enter a grade: "; cin >> grade; if (grade > 90) { cout << "excellent score" << endl; } else if (grade > 80) { cout << "good work, keep it up" << endl; } else if (grade > 70) { cout << "ok, keep working on the material" << endl; } else if (grade > 60) { cout << "be sure to ask questions when things are hard" << endl; } else { cout << "you need to work very hard" << endl; }

  Each else clause is necessary. If they're left out a grade of 100 would cause all the messages to be printed. In the code above a grade of 75 will cause the guard (grade > 70) to be true, and that's the only guard that will be evaluated as true.

• 4.8 The output of cout << (9 * 3 < 4 * 5) << endl; is 0 because the expression evaluates to 27 < 20 , which evaluates to false, which is 0. This is because the relational operator < has lower precedence than the arithmetic operators (see Howto A and Table A.4 on 715).

  The expression

  cout << (9 * (3 < 4) * 5) << endl; evaluates to 9 * 1 * 5 because (3 < 4) is true which is 1. This is why 45 is output. The parentheses are needed because < (less than); has lower precedence than * (multiplication).

• 4.9: The statement cout << 9 * 5 < 45 << endl generates 0 as output because it is false and false corresponds to 0 for output. If <= were used instead of < then 1 would be printed since the statement would be true.

  The expression (9*5 < 45 < 30) evaluates to 45 < 45 < 30 which evaluates to 0 < 30 which is 1 (true). This depends on the associativity of the < operator.

• 4.10: There are many possibilities, one is shown below. string grade; int score; cout << "enter score: "; cin >> score; if (score > 100) { cout << "score too high, between 0 and 100" << endl; } else if (score >= 80) { grade = "High Pass"; } else if (score >= 60) { grade = "Pass"; } else{ grade = "Fail"; }

• 4.11: Here's a program fragment for BTU's to Joules; note that 10,000/9.48 == 1054.8. double btu; cout << "enter BTU's "; cin >> btu; cout << btu << " BTU's = " << btu * 1054.8 << " Joules"; To use an assignment operator, use: double joule; double btu; cout << "enter BTU's "; cin >> btu; joule = btu * 1054.8; cout << btu << " BTU's = " << joule << " Joules";

  To use a function:

  double BtuToJoule(double btu) // post: returns number of joules in btu British Thermal Units { return btu * 1054.8; } To convert knots to miles per hour modify the program fragments above for BTUs but use different conversions. double knots; cout << "enter knots: "; cin >> knots; double mph = knots * 101.269 * 60 / 5280. The assignment to mph comes from canceling units as shown below

            101.269 feet     60 minutes     1 mile
            ------------  *  ----------  * ---------
             1  minute         1 hour        5,280 feet
  
  

• 4.12: The modified Cost is shown below double Cost(double radius, double price) // postcondition: returns the price per sq. inch { return price/(3.14159*pow(radius,2)); }

• 4.13: The code will never generate an error because of short-circuit evaluation. If the value of x is negative when if (x >= 0 && sqrt(x) > 10) is evaluated the subexpression x >= 0 will be false, and the call to sqrt will NOT be made because the whole expression must be false. This is one of the nice things about short-circuit evaluation of boolean expressions.

• 4.14: One version of function TriangleArea is shown below. There are other ways to write the function. double TriangleArea(double a, double b, double c) // precondition: a,b,c represent sides of a triangle // postcondition: returns area of triangle with sides a, b, c { double semip = (a + b + c)/2.0; return sqrt(semip * (semip - a) * (semip - b) * (semip - c)); }

• 4.15: One version of function SideLength is shown below. There are other ways to write the function. The use of the local variable c isn't necessary, the function could be one line: return sqrt(a*a + b*b - 2*a*b*cos(angle)) double SideLength(double a, double b, double angle) // precondition: a and b are lengths of sides of a triangle, angle // is the radian measure of angle between a and b // postcondition: returns length of side opposite a and b { double c = a*a + b * b - 2 * a * b * cos(angle); return sqrt(c); }

• 4.16: The average of three values can be output using the statement below. cout << (a + b + c)/3.0 << endl; The parentheses are necessary because / has higher precedence than +. If the type of a,b,c is changed to double then the average will be a double as well. However, in the statement above a double value will be calculated even if a,b,c are int's because of the 3.0. The operator >> is left associative. The statement cin >> a >> b >> c is treated like ((cin >> ) >> b) >> c .

• 4.17: There are many ways of writing this code. In the function below I've used a return statement on a single line with an if statement (for February). In some situations it's convenient not to use a compound statement. In such situations I put all the code on one line as shown below. in DaysInMonth(int month, int year) // precondition: month coded as 1 = january, ..., 12 = december // postcondition: returns # of days in month in year { // 30 days has september, april, june, and november if (9 == month || 4 == month || 6 == month ||11 == month) { return 30; } else if (2 == month) // handle february { if (IsLeap(year)) return 29; else return 28; } else { return 31; } }
• 4.18: The parentheses are needed because the expression TensPrefix(10 * num / 10) when num == 22 evaluates to TensPrefix(10 * 22 / 10) which is TensPrefix(220/10) which is TensPrefix(22) because * and / have equal precedence so in the absence of parentheses the order of evaluation is left-to-right.

• 4.19: This function can be written in many ways, I've shown two below. The second version uses the same ``logic'' as the first, but returns the result of the expression testing for divisibility by 2 which is either true or false depending on whether num is odd or even, respectively. bool IsEven(int num) // precondition: returns true if num is even, else returns false { if (num % 2 == 0) { return true; } else { return false; } } // alternate version bool IsEven(int num) // precondition: returns true if num is even, else returns false { return num % 2 == 0; } Which of these is preferable? As with many questions,the answer is "it depends". Students tend to write solutions like the first one, with an if/else statement. This is fine, it's easy to see the logic. Many students have trouble realizing that expressions have results that can be returned, e.g., the result of evaluating an == operator is a boolean value. Students should understand the second solution, but neither of these is qualitatively better than the other.

  It's ok to use % here negative numbers because although the value of (-3 % 2) may not be 1 --- it's -1 on most systems --- this value is NOT zero, so it's use in the code above will work since any negative even number will yield a remainder of zero when divided by 2, even in C++.

  If you want to avoid using the % (mod) operator, the code below can be used:

  bool IsEven(int num) // precondition: returns true if num is even, else returns false { int divbytwo = num/2; // truncates, so 3/2 = 1 return (divbytwo * 2 == num); }

• 4.20: In the function below I've used the one line form of an if statement again. In this case the code is clear and it probably won't be modified so this approach is justified. string DayName(int day) // precondition: 0 <= day <= 6 // postcondition: returns string representing day with // 0 = "Sunday", 1 = "Monday", ..., 6 = "Saturday" { if (0 == day) return "Sunday"; else if (1 == day) return "Monday"; else if (2 == day) return "Tuesday"; else if (3 == day) return "Wednesday"; else if (4 == day) return "Thursday"; else if (5 == day) return "Friday"; else if (6 == day) return "Saturday"; }

• 4.21: One relatively straightforward way to make Program 4.10, numtoeng.cpp work for three digit numbers is to rename the function NumToString to NumToString99 and use it only to handle numbers between 0 and 99 (inclusive). The function NumToString shown below will then call NumToString99 in converting all numbers, including three digit numbers, to English equivalents.
  string NumToString(int num) // precondition: 0 <= number <= 999 // postcondition: returns english equivalent, e.g., 101->one hundred one { if (0 <= num && num <= 99) { return NumToString99(num); } else // 100 <= num { string hundreds = DigitToString(num / 100) + " hundred "; if (num % 100 == 0) // 100, 200, 300,... 900 { return hundreds; } else { return hundreds + NumToString99(num % 100); } } }

• 4.22: This function is easier to write than the leap year function for our (Julian/Gregorian) calendar. bool IsIslamicLeapYear(int year) // postcondition: returns true if year is a leap year in the // Islamic calendar, else returns false { return ((11*y + 14) % 30) < 11; }

• 4.23: I used only 2 if statements in the function below by using the && operator. Some prefer to use if-else rather than just if even though the early return after each if makes the use of else unnecessary. int IslamicDaysInMonth(int month, int year) // postcondition: returns # of days in month/year { // odd numbered months have 30 days if (month % 2 == 1) return 30; if (12 == month && IsIslamicLeapYear(year)) return 30; return 29; // even numbered months have 29 days }

• 4.24: To determine if a given year is a leap year, create a Date object for february and check the number of days in the month: bool IsLeap(int year) // post: returns true if and only if year is a leap year { Date feb(2,1,year); return feb.DaysIn() == 29; }

• 4.25: Here's code to determine the day of a week #include <iostream> using namespace std; #include "date.h" int main() { int m,d,y; cout << "enter month day year "; cin >> m >> d >> y; Date theDate(m,d,y); cout << theDate << " falls on a " << theDate.DayName() << endl; return 0; }

• 4.26: My birthday in 2002 is shown below, your birthday will be different since you'll need to change the values used to construct the variable birthDay int main() { Date birthDay(4,8,2002); cout << birthDay << " falls on a " << birthDay.DayName() << endl; return 0; }