Can't fix logical problem, pset2 substitution

Question

I can't fix this logical problem. I have a check_alpha function which should return if all characters are alphabetic.

bool check_alpha(char key[])
{
    for (int i = 0, n = strlen(key); i < n; i++)
    {
        if (isalpha(key[i]))
        {
            printf("everything ok\n");
        }
        else {
            printf("not alpha.\n");
        }
    }
    return false;
}

And before that in my main function I have this condition

 else if (argc == 2 && !(check_alpha(argv[1])))
    {
        printf("Key must only contain alphabetic characters.\n");
        return 1;
    }

With current code and with argv of 26 alphabetic characters, I still get "Key must only contain alphabetic characters." If I change my logic, it still doesn't work properly, doesn't matter what I try. On the other hand, printf statements in my check_alpha function show me that it is working properly. Any help, what am I doing wrong? Ideally, if all 26 characters are letters, it should procede to check next condition...

Answer 1

Using multiple expressions in a logical condition like IF always complicates understanding, I would use separate checks, I would also not use a function to check if a character is alpha or not, for that we have the function isalpha () !!. On a positive note, it is good that you dare to experiment with functions. However be careful, let's see your code, the function check_alpha () takes as an argument a string and always returns FALSE, now let's see the IF condition written in English (well, the English that I handle ...):

"if the number of arguments is two and the !check_alpha () function is true, the body of the IF condition is executed"

Let's take a closer look at this, check_alpha () returns FALSE, so !chek_alpha() is true, so if we have two arguments the IF condition is true. I hope it was clear, and it will help you.