title: passing argument makes pointer from integer without a cast
body: <p>I've read through several similar questions on Stack Overflow, but I've not been able to find one that helps me understand this warning in this case. I'm in my first week of trying to learn C though, so apologies if I've missed an obvious answer elsewhere on Stack Overflow through lack of understanding.</p>
  
  <p>I get the following warning and note:</p>
  
  <pre><code> warning: passing argument 2 of ‘CheckIfIn’ makes pointer from integer without a cast [enabled by default]
   if(CheckIfIn(letter, *Vowels) ){
   ^
  
   note: expected ‘char *’ but argument is of type ‘char’
   int CheckIfIn(char ch, char *checkstring) {
  </code></pre>
  
  <p>When trying to compile this code:</p>
  
  <pre><code>#include &lt;stdio.h&gt;
  #include &lt;string.h&gt;
  #define CharSize 1 // in case running on other systems
  
  int CheckIfIn(char ch, char *checkstring) {
      int string_len = sizeof(*checkstring) / CharSize;
      int a = 0;
  
      for(a = 0; a &lt; string_len &amp;&amp; checkstring[a] != '\0'; a++ ){
  
          if (ch == checkstring[a]) {
              return 1;
          }
      }
      return 0;
  }
  
  
  // test function    
  int main(int argc, char *argv[]){
      char  letter = 'a';
      char *Vowels = "aeiou";     
  
      if(CheckIfIn(letter, *Vowels) ){
          printf("this is a vowel.\n");
      }
  
      return 0;
  }
  </code></pre>