title: Understanding the getting of task_struct pointer from process kernel stack
body: <p>Right now I'm reading the book &quot;Linux Kernel Development 3d Edition&quot; by Robert Love. There he write about the thread_info struct which contains the pointer to task_struct struct and, as I understood, located at the bottom or at the top of kernel stack of process (depends on architecture). I wasn't familiar with Linux kernel API until recently and I wasn't known of  current() method existence. There is an excerpt from the book related to how current() method actually works:</p>
  <blockquote>
  <p>On x86, current is calculated by masking out the 13 least-significant bits of the stack
  pointer to obtain the thread_info structure.This is done by the
  current_thread_info() function.The assembly is shown here:
  movl $-8192, %eax
  andl %esp, %eax
  This assumes that the stack size is 8KB.When 4KB stacks are enabled, 4096 is used in
  lieu of 8192.</p>
  </blockquote>
  <p>My questions are:</p>
  <ol>
  <li>As far as I know if we have a decimal value represented as a set of bits, then there is only one least-significant bit in the set, isn't it?</li>
  <li>What is the magical number 13?</li>
  </ol>
  <p>For thous who will read this topic, the questions I have voiced can lead to conclusion that the author don't understand properly the process of memory allocation and administration. Ok, that's may be right due to the fact that in my mind I can represent the memory allocated for the stack as the ribbon full of bits (or bytes). All of this bytes accessible by a specific memory address represented as some decimal value. The origin of the stack is the lowest memory address and the fin of the stack is the highest value of memory address. But HOW, HOW can we get the pointer to the thread_info struct located at the, say, end of the stack only by masking out 13 least-significant bits of ARBITRARY located stack pointer (If I understood correctly, we masking out bits of the stack pointer ADDRESS represented as decimal value).</p>