Question

Write an expression that continues to bid until the user enters 'n'.

#include
using namespace std;
int main() {
char keepGoing;
int nextBid;
nextBid = 0;
keepGoing = 'y';
while (/* Your solution goes here */) {
nextBid = nextBid + 3;
cout << "I'll bid $" << nextBid << "!" << endl;
cout << "Continue bidding? (y/n) ";
cin >> keepGoing;
}
cout << endl;
return 0;
}

Answer 1

The provided C++ code fragment is designed to prompt the user to enter 'y' to continue bidding, incrementing the bid amount by $3 each time. The correct condition for the while loop is keepGoing == 'y', which checks if the user wants to keep bidding and continues the loop accordingly.

Step-by-step explanation:

The student in question is attempting to implement a loop within a C++ program that will repeatedly ask the user if they want to continue bidding. The program increases the nextBid by $3 with each iteration of the loop, and the loop continues as long as the user enters 'y' (yes) when prompted to continue bidding. The correct condition to be placed inside the while loop is keepGoing == 'y', which checks if the keepGoing variable contains the character 'y'.The provided code will properly execute the program's intended functionality by using this condition. Here's how the loop condition should be written:while (keepGoing == 'y') {This will ensure that the loop will continue to iterate and an additional $3 bid will be added as long as the user's input is 'y'. When the user enters 'n', the loop will terminate, and the program will end.