SOLUTIONS TO EXERCISES PART ONE: Problem-Solving Techniques

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1 SOLUTIONS TO EXERCISES PART ONE: Problem-Solving Techniques 1 Principles of Programming and Software Engineering 1 const CENTS_PER_DOLLAR = 100; void computechange(int dollarcost, int centscost, int& d, int& c); // // Computes the change remaining from purchasing an item costing // dollarcost dollars and centscost cents with d dollars and c cents. // Preconditions: dollarcost, centscost, d and c are all nonnegative // integers and centscost and c are both less than CENTS_PER_DOLLAR. // Postconditions: d and c contain the computed remainder values in // dollars and cents respectively. If input value d < dollarcost, the // proper negative values for the amount owed in d dollars and/or c // cents is returned. // a const MONTHS_PER_YEAR = 12; const DAYS_PER_MONTH[] = 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31; void incrementdate(int& month, int& day, int& year); // Increments the input Date values (month, day, year) by one day. // Precondition: 1 <= month <= MONTHS_PER_YEAR, // 1 <= day <= DAYS_PER_MONTH[month - 1], except // when month == 2, day == 29 and isleapyear(year) is true. // Postcondition: The valid numeric values for the succeeding month, day, // and year are returned. bool isleapyear(int year ); // Determines if the input year is a leap year. // Precondition: year > 0. // Postconditions: Returns true if year is a leap year; false otherwise. 2b void incrementdate(int& month, int& day, int& year) // error check on input values if (month <= 0 month > MONTHS_PER_YEAR) cout << "Bad input value for month" << endl; if (year <= 0) cout << "Bad input value for year" << endl; Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley. 1

2 if (day <= 0 day > DAYS_PER_MONTH[month - 1]) // special case for February if ( month!= 2 ) cout << "Bad input value for day" << endl; else if (!isleapyear(year) day > 29) cout << "Bad input value for day" << endl; // end if // end if // increment the date day++; if (day > DAYS_PER_MONTH[month - 1]) // test for February 29 if (month == 2 && isleapyear(year) && day == 29) else // increment to next month day = 1; month++; if (month > MONTHS_PER_YEAR) // increment to next year month = 1; year++; // end if // end if // end if // end incrementdate bool isleapyear(int year) // Example: 2000 is a leap year but 1900 was not return ((year % 400 == 0) ((year % 4 == 0) && (year % 100!= 0))); // end isleapyear 3 We can make several improvements to user interaction and programming style: Prompt the user for the input and indicate the expected form of the input. The user should also be given an option to exit the program. Give more descriptive output. Make the program more robust; e.g., requiring exactly eight characters per name is unreasonable. Check input for obvious errors; e.g., a four-digit age entry is surely a typo and the user should be allowed to correct the error. Document the program. Use more descriptive variable names. Declare and use a constant for the maximum length of a name. 4 The code fragment is an infinite loop. The loop will never terminate because the condition will always be true. Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley. 2

3 5 There are two hazards with the function compute(). First, the argument X cannot be a negative number because compute() calls the sqrt() function with X as the argument. Second, since zero is in the range of the function cos(x), the result of this computation must be tested prior to using it as a divisor in order to avoid a "Divide by Zero error". The following implementation performs the required fail-safe checks. #include <cmath> #include <iostream> double compute(double x ) // // Performs a computation. // Preconditions: x >=0 and x!= (2k - 1)*PI/2 where k is a // positive integer. // Postcondition: returns -1 if preconditions are not met // else, returns the value of the computation. // // compute the cosine once and store it double denominator = cos(x); // check for negative argument to sqrt() function if (x < 0) cerr << "Cannot take square root of negative number" << endl; return -1; // check for division by zero if (denominator == 0) cerr << "Cannot divide by zero" << endl; return -1; return sqrt(x)/denominator; 6 The invariants for the for loop of the function factorial() are as follows: a 1 i n. b f = i! or n! 7 The loop invariant is as follows: 20 >= i >= 1 8 #include <iostream> int main () int n = 5; int square = 1; for (int i = 1; i <= n; i++) square = i * i; cout << square << endl; return 0; Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley. 3

4 9 The invariants for the for loop are as follows: a 0 i n - 1. b 0 numpos LIMIT. c sum is equal to the sum of the first numpos integers in anarray. const int LIMIT = 5; typedef int elementtype; elementtype computesum(const elementtype anarray[], int n, bool& success) - // Computes the sum of the first LIMIT positive elements in anarray. // Precondition: anarray is an array of n elementtypes with at least LIMIT // of anarray positive and n >= LIMIT. // Postcondition: If anarray contains at least LIMIT positive elements, then // the function returns the sum of the first LIMIT positive // elements and success is true. Else, sum is 0 and // success is false. anarray and n are unchanged. - int numpos = LIMIT; elementtype sum = 0; // check for out of bounds errors if (numpos > n n < 0) success = false; return sum; // end if // perform the computation on the array for (int i = 0; i < n && numpos > 0; ++i) if (anarray[i] > 0) sum += anarray[i]; numpos--; // end for // verify that LIMIT elements have been added if (numpos > 0) success = false; return 0; // end if // pass on results success = true; return sum; // end computesum 10 Invariant: 0 index n and sum = item[0] item[index]. Prior to the while loop, index is set to 0 and sum is initialized to item[index], i.e. item[0], and so the invariant describes the value of sum at this point. Within the while loop, the value of sum so far is item[0] item[index]. If index < n, index is then incremented (i.e.: index = index + 1). The value of the next element of the array with respect to the last accessed element (i.e.: item[ <last value of index> + 1 ]) is added to sum. Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley. 4

5 After the while loop, the last value of index must have been n since the loop terminated. Since index increments before it is used to access the array item, the last value of index (i.e.: n) must also have been used. Since the value of sum prior to this last access was item[0] item[n - 1] and the last quantity added was item[n], the algorithm corroborates the invariant at index = n. 11 The bug is in the while loop: a Output is The floor of the square root of 64 is 7. b The while loop condition should be while (temp1 <= x) Debugging involves printing the values of temp1, temp2 and ans at the top of the loop. c Supply user prompts and check the value of x to ensure that it is greater than #include <iostream> #include <cstdlib> enum errorcode DZERO, AOTRG,... ; // mnemonic error codes void severeerrormessage(errorcode error) // // Displays an error code and terminates program execution. // Preconditions: none. // Postconditions: An error message corresponding to the input // errorcode is output to the standard error stream // and the program is terminated. // switch(error) case DZERO: cerr << "Divide by zero error." << endl; break; case AOTRG: cerr << "Array index out of range." << endl; break;.. default: cerr << "Unknown fatal error." << endl; ; // end switch exit(0); // terminate execution immediately // end severeerrormessage Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley. 5