title: Why does paging let us use physical memory that is larger than what can be addressed by the CPU's address pointer length?
body: <p>I'm reading Operating System Concepts by Avi Silberschatz, in chapter 8 Main Memroy, it goes like:</p>
  
  <blockquote>
    <p>Frequently, on a 32-bit CPU, each page-table entry is 4 bytes long, but that
    size can vary as well. A 32-bit entry can point to one of 2<sup>32</sup> physical page frames.
    If frame size is 4 KB (2<sup>12</sup>), then a system with 4-byte entries can address 2<sup>44</sup> bytes
    (or 16 TB) of physical memory. We should note here that the size of physical
    memory in a paged memory system is different from the maximum logical size
    of a process. As we further explore paging, we introduce other information that
    must be kept in the page-table entries. That information reduces the number of bits available to address page frames.A 32-bit 
    CPU uses 32-bit addresses, meaning that a given process space can only be 2<sup>32</sup>
    bytes (4 GB). Therefore, paging lets us use physical memory that is larger than
    what can be addressed by the CPU’s address pointer length.</p>
  </blockquote>
  
  <p>I'm really confused with some concepts now. </p>
  
  <p>In the first example(4KB frame size and 2<sup>32</sup> frames), since we have 32-bit frame number and 12-bit offset, is the logical address 44-bit? So what's relationship between logical address space and the CPU's word size?</p>
  
  <p>In the second example(32-bit CPU using 32-bit addresses), how should I understand the physical memory that can be addressed by the CPU's address pointer length? And is the CPU's address pointer 32-bit long since CPU uses 32-bit addresses?</p>