P2_4_1.cpp - Read and average some integers, print the result. // This program continue asking for a new number until the user enters a 0 to terminate the program #include using namespace std; int main(void) { int x; int count = 0; // (1) initialize a counter to 0 to count number of values int choice = 1; // This is the choice that controls the looping continuation or termination double sum = 0; // initialize the sum to 0 to make sure the sum at the beginning is 0 double average;

while( choice == 1) // (2) read N grades and compute their sum, count ensures N entries { // read each number and compute the sum: cout << " Enter a grade : "; cin >> x; sum = sum + x; count++; // (3) update the count // prompt the user: cout << "Do you wish to enter another grade? (1 for yes and 0 or other key for no): " << endl; cin >> choice; }

if(count == 0) cout << "You haven't entered any number, no average will be computed, bye "; else{ average = sum/count; //Notice that we have divided by count this time cout << "The average of these " << count << " grades is " << average << endl; }

return 0; }

Exercise 2.4.1 In program P2_4_1.cpp, right at the beginning you have initialized the choice to 1. You did that to get the while loop to run at least once. If you use a do ... while instead, you wouldn't need to initialize the choice. Re-write the above program, call the new program ex2_4_1.cpp, so that is uses do ... while. Do not initialize the choice to 1 this time and compile and run the program. Does the program work the same way?