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;
    }
}