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    <p>Please look at this <em>very</em> carefully:</p> <pre><code>template &lt;class Type&gt; Type LinkList&lt;Type&gt;::GetItem(int itemNumber) const { Node&lt;Type&gt;* cursor = NULL; // loop over all items in the linked list for(cursor = first; cursor != NULL; cursor = (cursor -&gt; next)) { // for each item in the linked list, run a for-loop // counter from 0 to (used-1). for(int i = 0; i &lt; used; i++) { // if the passed in itemNumber matches 'i' anytime // before we reach the end of the for-loop, return // whatever the current cursor is. if(itemNumber == i) { return(cursor -&gt; data); } } } } </code></pre> <p>You're not walking the cursor down the list <code>itemNumber</code> times. The very first item <code>cursor</code> references will kick off the inner-for-loop. The moment that loop index reaches itemNumber you return. <strong>You never advance your cursor if the linked list has at least <code>itemNumber</code> items in the list.</strong>. In fact, the two of them (cursor and itemNumber) are entirely unrelated in your implementation of this function. And to really add irony, since <code>used</code> and <code>cursor</code> are entirely unrelated, if <code>used</code> is ever <strong>less</strong> than <code>itemNumber</code>, it will ALWAYS be so, since <code>used</code> doesn't change when <code>cursor</code> advances through the outer loop. Thus <code>cursor</code> eventually becomes NULL and the results of this function are undefined (no return value). In summary, as written you will always either return the first item (if <code>itemNumber &lt; used</code>), or undefined behavior since you have no return value.</p> <p>I believe you need something like the following instead:</p> <pre><code>template&lt; class Type &gt; Type LinkList&lt;Type&gt;::GetItem(int itemNumber) const { const Node&lt;Type&gt;* cursor = first; while (cursor &amp;&amp; itemNumber-- &gt; 0) cursor = cursor-&gt;next; if (cursor) return cursor-&gt;data; // note: this is here because you're definition is to return // an item COPY. This case would be better off returning a // `const Type*` instead, which would at least allow you to // communicate to the caller that there was no item at the // proposed index (because the list is undersized) and return // NULL, which the caller could check. return Type(); } </code></pre>
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    1. COThis is what happens when you code too late at night. I was so concerned with connecting and comparing the function outputs in int main that I forgot to really think about the relationship between cursor and the input itemNumber. Guess I just had For loops on the brain last night. Thanks again.
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    2. CO@DanteoftheInferno No kidding. Been down that road many times. Seriously consider what I mention in the proposal about returning a pointer to the item, and the caller checking it for NULL. Alternatively you could have it throw an OO-range exception and change the function to return a const Type&. Either way, don't make a copy unless that is *really* what you want to do. It can get VERY expensive for complex objects used as Type.
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