Reading digits and converting to words - c++

Question

I'm new to c++ and I have to write a program that takes a user 4-digit number and convert it to words i.e. 7238 would be wrote as seven two three eight. Yet it writes every number as unknown. Any advice for a noob would be greatly appreciated.
#include iostream
using namespace std;
int main() {
char number;
cout << "Please enter a 4 digit number: ";
cin >> number;
switch(number){
case 1 :
cout<< "one";
break;
case 2 :
cout<< "two";
break;
case 3 :
cout<< "three";
break;
case 4 :
cout<< "four";
break;
case 5 :
cout<< "five";
break;
case 6 :
cout<< "six";
break;
case 7 :
cout<< "seven";
break;
case 8 :
cout<< "eight";
break;
case 9 :
cout<< "nine";
break;
case 0 :
cout<< "zero";
break;
default :
cout << "UNKNOWN.";
}
}

Answer 1

Sounds like homework but here are some tips. Change your number variable to type of int You can break the number out into individual variables with division and modulus. I would stuff those into an integer array.
int array[4];
arr[0] = (number / 1000) % 10; // Thousands
....... // You do the hundreds and tens
arr[3] = (number % 10); // Ones
Then use a loop around your switch statement where your counter is less than 4 (the length of the array). Make sure to increase your counter at the end of each loop. Oh, and it's #include <iostream>.

Answer 2

With to_string and range based for:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int number;
cout << "Enter the number: ";
cin >> number;
string strnum = to_string(number);
for (auto c : strnum)
{
switch (c)
{
case '0': cout << "zero "; break;
case '1': cout << "one "; break;
case '2': cout << "two "; break;
case '3': cout << "three "; break;
case '4': cout << "four "; break;
case '5': cout << "five "; break;
case '6': cout << "six "; break;
case '7': cout << "seven "; break;
case '8': cout << "eight "; break;
case '9': cout << "nine "; break;
default: cout << "non-digit"; break;
}
}
return 0;
}

Answer 3

You need to put ascii values in your case statements. Currently you are comparing the ascii values for digits with numbers 0 - 9.
Values can be found here : http://www.asciitable.com/

Answer 4

Your variable is of type char. A char stores a character, usually ASCII encoded. If the user inputs a '1', for example, that would usually translate to an integer value of 49, not 1. Either read into an int or change your case labels to use character literals:
case '1':
cout << "one";
break;
You could then use a loop to read multiple digits.