C program to convert digit to words

CServer Side ProgrammingProgramming

Suppose we have a digit d, we shall have to convert it into words. So if d = 5, our output should be "Five". 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 = 6, then the output will be "Six".

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 <stdio.h>
void solve(int d){
    if(d < 0 && d > 9){
        printf("Beyond range of 0 - 9");
    }else if(d == 0){
        printf("Zero");
    }else if(d == 1){
        printf("One");
    }else if(d == 2){
        printf("Two");
    }else if(d == 3){
        printf("Three");
    }else if(d == 4){
        printf("Four");
    }else if(d == 5){
        printf("Five");
    }else if(d == 6){
        printf("Six");
    }else if(d == 7){
        printf("Seven");
    }else if(d == 8){
        printf("Eight");
    }else if(d == 9){
        printf("Nine");
    }
}
int main(){
    int d = 6;
   
    solve(d);
}

Input

6

Output

Six
raja
Published on 08-Oct-2021 10:56:42
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