title: X86_64 - assembly - Why displacement not 64 bits?
body: <p>I am reading the <em>Intel x86_64 guide vol.1</em> to refresh how memory addressing works.</p>
  
  <p>Still, </p>
  
  <blockquote>
    <p>3.7.5 Specifying an Offset</p>
    
    <p>The offset part of a memory address can be specified directly as a static value (called a displacement) or through an address computation made up of one or more of the following components:</p>
    
    <p>• Displacement — An 8-, 16-, or 32-bit value.</p>
  </blockquote>
  
  <p>I read in <em>Agner Fog's assembly guide</em> that 64-bit absolute addressing was possible when used with (r/e)ax register.</p>
  
  <p>So.. </p>
  
  <p><strong>Is it possible, or not, to use absolute addressing with 64 bits addresses to jmp, mov and call (with all the registers), or will I have to keep using the Base + displacement combo</strong>?</p>