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C++ program to convert digits to words using conditional statements
Suppose we have a digit d, we shall have to convert it into words. So if d = 9, our output should be "Nine". If we provide some d which is beyond the range of 0 and 9, it will return appropriate output.
So, if the input is like d = 3, then the output will be "Three".
To solve this, we will follow these steps −
- Define a function solve(), this will take d,
- 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")
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 d;
cin >> d;
solve(d);
}
Input
3
Output
Three
Updated on: 07-Oct-2021
702 Views
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