CODEWITHSUNDEEP
c++user-input
Sam LaManna
Sam LaManna

Reputation: 425

How to get user input in do/while loop?

I tried to use the do / while loop I asked about and fixed in my one function in int main to allow the entire program to be rerun if the user wants to, but it is rerunning the program without waiting for user input.

int main()
{
    int spoolnumber = 0;     // Number of spools to be ordered
    float subtotalspool = 0; // Spool sub total
    float shippingcost = 0;  // Shipping cost
    float totalcost = 0;     // Total cost
    char type = 'n';

    do {
        instruct();                     // Print instructions to user
        spoolnumber = spoolnum();       // calculate and store number of spools

        subtotalspool = stotalspool(spoolnumber);       // Calculate subtotal
        shippingcost = shipcost(subtotalspool);         // Calculate subtotal
        totalcost = tcost(subtotalspool, shippingcost); // Calculate final total

        // Print final output
        results(spoolnumber, subtotalspool, shippingcost, totalcost);     
        cout << "\n" << " Would you like to run the program again? [y/n]"; 
    }
    while (type != 'y');

    return 0;
}

Upvotes: 1

Views: 6690

Answers (5)

Xeo
Xeo

Reputation: 131799

Well, you never ask for input, do you? Add the following after the cout line:

cin >> type;
cin.ignore(); // remove trailing newline token from pressing [Enter]

Now, you'll still need the usual testing, if the input was valid at all etc, but this should get you going.

Upvotes: 6

Flexo - Save the data dump
Flexo - Save the data dump

Reputation: 88711

You haven't read any input from the user. You could simply do:

cin >> type;

But really you want to be checking that it is succeeding too, e.g. not eof or other errors otherwise it's still possible to loop forever if the user pressed Crtl-D for example.

Checking if it succeeds:

if (!(cin >> type)) {
   // Reading failed
   cerr << "Failed to read input" << endl;
   return -1;
}

Which you could actually make part of the loop condition:

while (cin >> type && type != 'y');

The advice from Xeo about calling cin.ignore() is important to since you will almost certainly end up with more than just one char worth of input.

Upvotes: 8

Nate
Nate

Reputation: 12829

That's because you're not prompting the user for input.

You'll be more successful if you try something like:

cout << "\n" << " Would you like to run the program again? [y/n]"; 
cin >> type;

Upvotes: 2

user195488
user195488

Reputation:

You need to include a cin in order to retrieve user's decision:

For example:

cout << "\n" << " Would you like to run the program again? [y/n]"; 
cin >> someVariable;

Upvotes: 2

Robᵩ
Robᵩ

Reputation: 168626

You haven't added any code to accept user input. At the bottom of your loop, trying reading a character from cin into type.

Also, you may need to flush the output of cout first, before accepting the user input from cin.

Upvotes: 8

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