Data Validation in C

Question

I was trying out a simple program in C for validating user data. The program is supposed to identify whether a user entered character is a number, alphabet or a special character.

Somehow , the code identifies every kind of input character as a number. I have appended the code below, I'd be grateful if someone could kindly point out where I'm going wrong ?

//Program to take input from the user and determine whether it is character, number, or a special character

#include<stdio.h>
#include<conio.h>
#include<string.h>


    char ch;

int main()

{
     clrscr();

    printf("Enter a character \n");
    scanf("%c \n",ch);

    if ((ch>='A'&& ch<='Z')||(ch>='a'&& ch<='z') )
    {
        printf("The character entered is an alphabet \n" );

    }
     else if ((ch>=0)&&(ch<=9))
    {
        printf("Character entered is an number \n");
    }


    else
    {
        printf("Character entered is a special character");

    }
    return 0;
}

Answer 1

scanf accepts a pointer as the argument for %c. In other words,

scanf("%c \n",ch);

should be written as:

scanf("%c\n",&ch);

Without the reference operator (&), scanf receives the value of ch. In this case, the value is garbage, because ch is unset.* Referencing ch gives scanf a pointer to ch, not ch itself, so scanf can modify the value of ch by dereferencing the pointer (using the dereference operator, *).

There's also the issue with digit checking that Himadri mentioned.

* This is actually undefined behaviour.

Answer 2

Oh, Arun very silly mistake. In your second condition in else if you have to right 0 and 9 in single quotation mark.

So, your code will be -

if ((ch>='A'&& ch<='Z')||(ch>='a'&& ch<='z') )
{
    printf("The character entered is an alphabet \n" );

}
 else if ((ch>='0')&&(ch<='9'))
{
    printf("Character entered is an number \n");
}
else
{
    printf("Character entered is a special character");

}

May be this is the only mistake. Now, it should work.