title: a pointer about *argv[]
body: <p>This is my <code>main.c</code> </p>
  
  <pre><code>......
  int main(int argc, char **argv)
  {
      init_arg(&amp;argc, &amp;argv);
      ......
  }
  </code></pre>
  
  <p>This is my <code>init_arg.c</code></p>
  
  <pre><code>......
  void init_arg(int *argc, char ***argv)
  {
      printf("%s\n", *argv[1]);
      ......
  }
  </code></pre>
  
  <p>I compiler it with no error and warning.</p>
  
  <p>I run it:</p>
  
  <p><code>./a.out include</code></p>
  
  <p>It get <strong>Segmentation fault</strong></p>
  
  <p>When I debug it， I found step  <code>printf("%s\n", *argv[1]);</code></p>
  
  <p>get wrong, It show:</p>
  
  <p><code>print *argv[1]</code></p>
  
  <p><code>Cannot access memory at address 0x300402bfd</code></p>
  
  <p>I want to know, How to print <code>argv[1]</code> in <code>init_arg()</code> function.</p>