here is the question posed.

# Ask the user to enter a digit between 0 and 9. Have the program print out the digit in words, for example:

Enter a digit between 0 and 9: 4

You entered the number four

Assume that the user will enter only a single digit. The user may accidentally enter a single character, and this should generate an error message.

when i type in other integers greater then 9 i display the error message, however when i type a character i return the your number is two.Code:#include <stdio.h> int main() { int num; printf("Please enter a digit from 0 through 9\n"); scanf("%d,",&num); if (num == 0) printf("The number you have typed is zero.\n"); else if (num == 1) printf("The number you have typed is one.\n"); else if (num == 2) printf("The number you have typed is two.\n"); else if (num == 3) printf("The number you have typed is three.\n"); else if (num == 4) printf("The number you have typed is four.\n"); else if (num == 5) printf("The number you have typed is five.\n"); else if (num == 6) printf("The number you have typed is six.\n"); else if (num == 7) printf("The number you have typed is seven.\n"); else if (num == 8) printf("The number you have typed is eight.\n"); else if (num == 9) printf("The number you have typed is nine.\n"); else printf("You have selected to not follow the instructions therefore an error has occurred.\n"); return 0; }

let me know where i went wrong.