In C, How can I check an array for only ASCII defined Letters Without using tons of If Statements?

I’m trying to make sure that the array entered at the command line is only letters, Capital and lowercase.

If it is numbers or anything else then I want to end the program.

Now, I know I can do this with tons of if statements to see:

if it is between this and that

else this and that

else this and that.

But I would like to try and learn a more efficient way to do this because I’m sure there is one with a for loop I just havent figured it out. I ask because I want the user to enter a key that is x amount long. And to achieve the desired / required string length I used a similar method of multiple if else statements.

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


int validate_key(int argc, string argv[]);

int main(int argc, string argv[])
{
    string input;
    int score;

    score = validate_key(argc, argv);

    if (score == 0)
    {
       // if required key is valid, next function of program here

       
    }

}

int validate_key(int argc, string argv[])
{

    //KEY VALIDATION 
    int l = 0; //initialize length variable
    string s;

    //To make sure that the string entered at command line is the required 26 characters, 
   //and that it isnt empty

    if (argc == 2) //make sure there are no spaces in the entered string
    {
        s = argv[1];     //assign key to string s
        l = strlen(s);   //check length of KEY

    }

    //check to see string isnt NULL / empty
    if (argc !=0 && l == 26) 
    {


        //make a for loop that scans the array making sure it is between two values? 
       //attempting to check argv to make sure it is a lowercase and or uppercase string 

         for (int i = 0; i < l; i++)
            {
                int charValue = argv[1][i];

                if ( ( charValue > 64 && charValue < 91 ) || ( charValue > 96 && charValue < 123 ) )
                {

                    printf("%c\n", charValue); //to show me that it made it into the if statement
                }
                else

                return 1;
            }


        return 0;

    }
    else if (l <= 25 && l >= 1) //entered key is too small
    {
        printf("Please Enter 26 Letter Key \n");
        printf("value of l: %d\n", l);

        return 1;
    }

    else //entered key is empty
    {
       //printf("value of l: %d\n", l);

        printf("missing command-line argument\n");

        return 1;

    }

}

>Solution :

Use isalpha():

7.4.1.2 The isalpha function

Synopsis

      #include <ctype.h>
      int isalpha(int c);

Description

The isalpha function tests for any character for which isupper or
islower is true, or any character that is one of a locale-specific set
of alphabetic characters for which none of iscntrl, isdigit, ispunct,
or isspace is true.200) In the "C" locale, isalpha returns true only
for the characters for which isupper or islower is true.

In your code:

     for (int i = 0; i < l; i++)
        {
            // see notes below regarding the cast
            int charValue = ( unsigned char ) argv[1][i];

            if ( isalpha( charValue ) )
            {

                printf("%c\n", charValue); //to show me that it made it into the if statement
            }
            else

            return 1;
        }

Note, though, that the value passed must be representable as an unsigned char:

The header <ctype.h> declares several functions useful for classifying and mapping characters. In all cases the argument is an int, the value of which shall be representable as an unsigned char or shall equal the value of the macro EOF. If the argument has any other value, the behavior is undefined.