CSCI 159 Quiz 2 Sample Solutions Monday lab

Question 1 [7 marks]
The C++ function below uses a while loop to achieve its repetition. Rewrite the function so that it uses recursion instead (no loops permitted). The behaviour from a user perspective must be as close to the original as possible.

// displays the integer values from high down to low (inclusive)
// with a "Done" message at the end
void dispVals(int low, int high)
{
   while (low <= high) {
      cout << high << endl;
      high--;
   }
   cout << "Done!" << endl;
}

Sample solution

void dispVals(int low, int high)
{
   if (low < = high) {
      cout << high << endl;  // get current (highest) value first
      dispVals(low, high-1); // then recurse on lower values
   } else {
      cout << "Done!" << endl; // must be in else to ensure it only happens in last call
   }
}

Question 2 [7 marks]
Below is a prototype for function getNegative and comments describing its desired behaviour. Implement (write the code for) the function, using a loop to achieve the repetition (no recursion permitted).

// the function prompts the user to enter a number greater than 1,
//    performs appropriate error checking and gets the user
//    to try again until a valid response is provided,
// and returns the eventual valid response
float getAbove10();

Sample solution

float getAbove1()
{
   float num;
   bool isOK;
   do {
      isOK = false;
      cout << "Enter a number greater than or equal to 1" << endl;
      cin >> num;
      if (cin.fail()) {
        cin.clear();
        cin.ignore(80,'\n');
        cerr << "Your input must be a number, try again" << endl;
      } else if (num <= 1) {
        cerr << "The value must be > 1, try again" << endl;
      } else {
         isOK = true;
      }
   } while (!isOK);
   return num;
}

Question 3 [6 marks]
Assuming the getANumber function correctly gets and returns a number from the user:
(i) Explain the conditions under which the code segment below results in an 'infinite loop' and suggest a simple fix.
(ii) Explain the conditions under which the body of the loop might never execute.

double num1, num2;
num1 = getANumber();
num2 = getANumber();
while (num1 < num2) {
   cout << num1 << endl;
   num1 = num1 * 2;
}
cout << "Done!" << endl;

Sample solution

(i) If num1 < = 0 and num2 > num1 then we can double num1 as often
    as we like and it'll still be < num2 (at least until it exceeds
    the biggest negative number a double can hold).

    One possible fix that doesn't alter the functionality for any other
    cases is to expand the while loop test case to something like
       while ((num1 > 0) && (num1 < num2)) {

(ii) The loop body never runs if num1 >= num2 from the very start.