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copied!<p>It's not the <code>push cs</code> that ensures this, it's the <code>push cs; pop ds;</code> combination that does.</p> <p>The first instruction copies the current value of <code>cs</code> onto the stack, and the second pulls that value off the stack and puts it into the <code>ds</code> register.</p> <hr> <p>In response to your request for more information, let's start with the following stack and registers:</p> <pre><code>stack=[1,2,3], cs=7, ds=6 </code></pre> <p>After <code>push cs</code>, which pushes the value of the <code>cs</code> register onto the stack:</p> <pre><code>stack=[1,2,3,7], cs=7, ds=6 </code></pre> <p>After <code>pop ds</code>, which pops a value off the stack and put it into the <code>ds</code> register:</p> <pre><code>stack=[1,2,3], cs=7, ds=7 </code></pre> <p>And that's basically it.</p> <hr> <p>I can't recall of the top of my head whether it was possible to transfer between segment registers with a <code>mov</code> instruction (I don't think it was, but I may be wrong, and this would necessitate the push/pop sequence). This <a href="http://www.penguin.cz/~literakl/intel/m.html#MOV" rel="nofollow">link</a> would seem to confirm that: there is no <code>mov</code> option with a segment register as both source <em>and</em> destination.</p> <p>But even if it were, assembler coders often chose more suitable instructions, either for speed or compact code (or both), things like using <code>xor ax, ax</code> instead of <code>mov ax, 0</code> for example.</p>

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