title: warning: format ‘%p’ expects argument of type ‘void *’, but argument 3 has type ‘char **’
body: <p>I'm trying to print out the value of <code>argv</code> at this point in my code, but when I try to compile it, I get
  <code>warning: format ‘%p’ expects argument of type ‘void *’, but argument 3 has type ‘char **</code>. I'm not sure what this warning means though. Is <code>%p</code> not meant to be used for <code>argv</code> even though it's a pointer?</p>
  
  <pre><code>int main(int argc, char* argv[])
  {
      printf("%s%p", "argv = ", argv);
  }
  </code></pre>