title: Difference between declaring shellcode as a char[] array and char*?
body: <p>Hi all,</p>
  
  <p>I'm trying to learn basic shellcoding and I've run across something curious that I hope someone can explain to me. I've compiled the following code two ways: declaring the shellcode as an array and as a char*. When I declare shellcode as an array, linux detects that I am trying to execute data and I get a segfault on the first instruction. However, when I declare shellcode as a char* all of the shellcode executes and I get a "Hello world!". How does the compiler treat these two declarations differently and why does one end in shellcode that lives in memory that is unprotected? Thanks in advance.</p>
  
  <pre><code>#include &lt;stdio.h&gt;
  #include &lt;stdlib.h&gt;
  #include &lt;string.h&gt;
  
  /* This declaration ends in a segfault */
  //char shellcode[] =
  
  /* This declaration ends in successful execution */
  char* shellcode = 
  
  /* Shellcode prints "Hello world!" and exits */    
  "\xeb\x1f\x48\x31\xc0\x48\x31\xdb\x48\x31\xc9\x48\x31\xd2\xb0\x04\xb3\x01\x59\xb2\x0c\xcd\x80\x48\x31\xc0\xb0\x01\x48\x31\xdb\xcd\x80\xe8\xdc\xff\xff\xff\x48\x65\x6c\x6c\x6f\x20\x57\x6f\x72\x6c\x64\x21";
  
  int main()
  {
      void (*f)();
      f = (void (*)())shellcode;
      (void)(*f)();
  }
  </code></pre>