I need help writing this option in ASSEMBLY code it takes in a string and removes all non letter characters then displays the new string using
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<pre class="ql-syntax">I need <span class="hljs-keyword">help</span> writing

this <span class="hljs-keyword">option</span> <span class="hljs-keyword">in</span> <span class="hljs-keyword">ASSEMBLY</span> code it takes <span class="hljs-keyword">in</span> a <span class="hljs-keyword">string</span> <span class="hljs-keyword">and</span> removes all non letter <span class="hljs-keyword">characters</span> <span class="hljs-keyword">then</span> displays the <span class="hljs-keyword">new</span> <span class="hljs-keyword">string</span> <span class="hljs-keyword">using</span> another <span class="hljs-keyword">procedure</span> that uses writestring <span class="hljs-keyword">and</span> <span class="hljs-keyword">register</span> edx ;// <span class="hljs-comment">------------------------------------------------------------------------------</span> option3 PROC ;// Description: removes all non-letter elements. There is no requirement for ;// option2 to have been executed. ;// Receives: ecx - length of string ;// edx - offset of string ;// ebx - offset of string length variable ;// esi preserved ;// Returns: nothing, but the string will have all non-letter elements removed .data .code push esi <span class="hljs-keyword">call</span> option5 L3: mov al, <span class="hljs-keyword">byte</span> ptr [edx+esi] cmp al, <span class="hljs-number">41</span>h jb notletter cmp al, <span class="hljs-number">5</span>Ah ja notletter cmp al, <span class="hljs-number">61</span>h jb notletter cmp al,<span class="hljs-number">7</span>Ah ja notletter mov <span class="hljs-keyword">byte</span> ptr [edx+esi], al notletter: inc esi <span class="hljs-keyword">loop</span> L3 pop esi <span class="hljs-keyword">call</span> option5 <span class="hljs-keyword">call</span> waitmsg option3 ENDP </pre>

Top Answer

The pseudocode will be i= 0 j= 0 while i&lt;len(str) and j&lt;len(str) if str[j] is letter str[i]=str[j] i=i+... View the full answer

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