Question

Im trying to add code so if the user mistypes they can go back in the program and retype the input but im not sure if the code i found and used is correct. here it is in my function:

/********************************************/
// Name: inspools                            /
// Description: Ask for and get number of    /
// spools                                    /
// Parameters: N/A                           /
// Reture Value: spoolnum                    /
/********************************************/
int spoolnum()
{
  int spoolnum;
  char type;

  cout << "Number of spools to be shipped: " << endl;
  cin >> spoolnum;
  cout << spoolnum << " spool(s) of wire will be shipped" << endl;
  cout << "Is this correct? [y/n] ";
  cin >> type;
  if ('n') << spoolnum;

  if ('y') break;

  return spoolnum ;
}

Answer 1

You said you searched for loops, but I don’t buy it. I imagine you are pretty new at programming. I’m going to give you the answer but not without some explanation first.

How While Loops Work

From Wikipedia:

In most computer programming languages, a while loop is a control flow
statement that allows code to be executed repeatedly based on a given
boolean condition. The while loop can be thought of as a repeating if
statement.

Your Problem

Your problem is that you want to keep making the user enter a choice until they enter y. To do this, you need at least a WHILE loop, or as other commenters have said a DO/WHILE loop.

I have never preferred DO/WHILE loops but others do prefer it.

The problems you may have with the below code is that you have more than just y returned in cin such as a newline (\n) character. You will have to handle that condition.

int spoolnum()
{
  int spoolnum = 0;
  char type = 'n';

  while (type != 'y') {
      cout << "Number of spools to be shipped: " << endl;
      cin >> spoolnum;
      cout << spoolnum << " spool(s) of wire will be shipped" << endl;
      cout << "Is this correct? [y/n] ";
      cin >> type;
  }
  return spoolnum;
}

or the alternative DO/WHILE:

int spoolnum()
{
  int spoolnum = 0;
  char type = 'n';

  do {
      cout << "Number of spools to be shipped: " << endl;
      cin >> spoolnum;
      cout << spoolnum << " spool(s) of wire will be shipped" << endl;
      cout << "Is this correct? [y/n] ";
      cin >> type;
  } while (type != 'y');

  return spoolnum;
}

In the above code, I removed your if ('n') << spoolnum; because frankly it does not make sense.

I also removed if ('y') break; because the while(...) loop will break once the condition is met, which is type equal to 'y'.