c - Checking if Char contains numbers and converting it to INT -

This program is doing well, but I am trying to check it, if user entered Value is the number of letters or not signs) So basically I am using ifdigit or '/ 0' with condition. The program is compiled properly, but I'm getting the error when entering less than 3 types (even numbers ...) what's wrong in it? I need only one thing when the number is 0-999 and when it is not ask again, or this number is not, then loop is to be left.

  char x [5]; Bool ijud; Int tab [20] [2], I = 0, A; Do {isgood = TRUE; Printf ("Write a where 0  

Incorrect:

< Li> Your Boolean-Logic
• Your Blank-Character
• Calling atoi when isgood is FALSE Change it:

    isdigit (x [j]) == 0 || X [j]! = '/ 0'  

  To do this:

    isdigit (x [j]) == 0 and ; Amp; X [j]! = '\ 0'  

  And this:

    a = atoi (x); If break (a & lt; 1000 & isgood == true);  

  To do this:

    if (isgood == true) {a = atoi (x); Break (0 & lt; a & amp; l & gt; 1000); }  

  There is a very easy way to do the whole thing:

    int x; Four c; Do {printf ("Type where 0