Repeating Input Requests

Question

I'm trying to code a simple program that prints out stars in the shape of a hill. It works by taking a user input width and printing the number of stars up until the width specified is reached. The main section of code that's printed is finished. However, I'm having issues trying to ensure the inputted number is odd. If the number is even, the function is supposed to continue asking the user to input another number until they input an odd number.

The relevant part of the code:

int rowsinput;

cout << "Enter an odd number width: ";
cin >> rowsinput;

if (rowsinput % 2 == 0)
{
    cout << "Please enter an odd number width: ";
}
else

If anyone needs me to post the entire code please comment.

Luis Gonzalo Quijada Romero · Accepted Answer · 2015-10-07 01:54:27Z

1

you can use something like:

int rowsinput;
do{
  cout << "Enter an odd number width: ";
  cin >> rowsinput; 
}while(rowsinput % 2 == 0)

answered Oct 7, 2015 at 1:54

Luis Gonzalo Quijada Romero

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1 Comment

user4581301 Over a year ago

Potential infinite loop with input "qwerty" and allows for rowsinput to be used uninitialized.

AkaSh · Accepted Answer · 2015-10-07 02:14:16Z

0

Add cin statement inside while loop as follows:

while(rowsinput % 2 ==0)
{
    cout << "Please enter an odd number width: ";
    cin >> rowsinput;
}

Your final code will be as follows :

cout << "Enter an odd number width: ";
cin >> rowsinput;
while(rowsinput % 2 ==0)
{
    cout << "Please enter an odd number width: ";
    cin >> rowsinput;
}

edited Oct 7, 2015 at 2:14

answered Oct 7, 2015 at 1:58

AkaSh

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1 Comment

user4581301 Over a year ago

And what of a second invalid input?

user4581301 · Accepted Answer · 2015-10-07 02:39:31Z

0

Consider as a substitute

while (!(cin >> rowsinput) ||  // bad, non-parsable input
       (rowsinput % 2 == 0) || // even number
       (rowsinput  < 0))       // negative number
{
    cout << "Please enter an odd number width: ";
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    cin.clear();
}
// input is good. Carry on with rest of program

This will loop until the user provides a valid integer, that integer is odd, and the integer is positive. If any of those conditions is not met, the user is prompted for more input, all current (bad) input is discarded, and any error conditions on the input stream are cleared.

Edit:

Requires an additional #include <limits> to get the numeric limits in the ignore call.

edited Oct 7, 2015 at 2:39

answered Oct 7, 2015 at 1:57

user4581301

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