I am currently trying to make my output display but I can't get it working
I would like it to display like this:
***********
*--------------*//ignore the - signs as I had to use them to hold position of *
* 5! is: 120 *
*--------------*//ignore the - signs as I had to use them to hold position of *
***********
and I am trying to do this as a function. Here is my code but it won't compile, I think it has something to do with using int value in the string "message"
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//number is entered by user and is of type int and it is this number my program calculates the factoral for.
//factoral is of type int and is the value of the factoral of the number entered
void print_factoral (int number, int factoral)
{
const string message = "" + number + "! is: " + factoral + "";
const string spaces (message.size(), " ");//used to build the 2nd and 4th line of output
const string second = "* " + spaces + " *";//creates 2nd and fourth line of output
const string first (second.size(), "*");//creates the 1st and 5th line of output
cout << endl;
cout << first << endl;
cout << second << endl;
cout << "* " << message << " *" << endl;
cout << second << endl;
cout << first << endl;
}
void print_factoral (int number, int factoral)
{
ostringstream oss;//creating a stringstream
oss << "" << number << "! is " << factoral << "";//adding display message to stringstream
const string message = oss.str();//assigning stringstream to message
const string spaces (message.size(), " ");//used to build the 2nd and 4th line of output
const string second = "* " + spaces + " *";//creates 2nd and fourth line of output
const string first (second.size(), "*");//creates the 1st and 5th line of output
cout << endl;
cout << first << endl;
cout << second << endl;
cout << "* " << message << " *" << endl;
cout << second << endl;
cout << first << endl;
}
It isn't working for me but I have no experience of using stringstream so not sure if I have done it correctly.
The program isn't compiling and I am getting an error saying invalid conversion from const char to char.
I am getting this message whether or not I use string or const string.