C++ program to convert all digits from the given range into words

Suppose we have two digits a and b. We shall have to convert each digit into words and print them one by one. Printing digits into words means for a digit 5, it should print "Five".

So, if the input is like a = 2, b = 6, then the output will be

Two Three Four Five Six

To solve this, we will follow these steps −

• if d < 0 and d > 9, then:
  • return ("Beyond range of 0 - 9")
• otherwise when d is same as 0, then:
  • return ("Zero")
• otherwise when d is same as 1, then:
  • return ("One")
• otherwise when d is same as 2, then:
  • return ("Two")
• otherwise when d is same as 3, then:
  • return ("Three")
• otherwise when d is same as 4, then:
  • return ("Four")
• otherwise when d is same as 5, then:
  • return ("Five")
• otherwise when d is same as 6, then:
  • return ("Six")
• otherwise when d is same as 7, then:
  • return ("Seven")
• otherwise when d is same as 8, then:
  • return ("Eight")
• otherwise when d is same as 9, then:
  • return ("Nine")
• From the main method, do the following:
• for i in range a to be, do
  • solve(i)
  • move cursor to the next line

Example

Let us see the following implementation to get better understanding −

#include <iostream> using namespace std; void solve(int d){     if(d < 0 || d > 9){         cout << "Beyond range of 0 - 9";     }else if(d == 0){         cout << "Zero";     }else if(d == 1){         cout << "One";     }else if(d == 2){         cout << "Two";     }else if(d == 3){         cout << "Three";     }else if(d == 4){         cout << "Four";     }else if(d == 5){         cout << "Five";     }else if(d == 6){         cout << "Six";     }else if(d == 7){         cout << "Seven";     }else if(d == 8){         cout << "Eight";     }else if(d == 9){         cout << "Nine";     } } int main(){    int a = 2, b = 6;     for(int i = a; i <= b; i++){         solve(i);         cout << endl;     } } 

Input

2, 6

Output

Two Three Four Five Six