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POC++ subscription operator overload
 

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    <p>I'm working on brushing up on my C++ during some project downtime and have created a a Linked List project to do so. In this project I would like to return the current value at a set index. I've already got a method to do this, but want to work it out with an operator overload.</p> <p>To that end I did the necessary research to brush up and have created the following code (I won't subject everyone to my code and will only paste the pertinent sections):</p> <pre><code>T&amp; operator[](const int index); template&lt;class T&gt; T&amp; LinkedList&lt;T&gt;::operator[](const int index) { try { if(!isIndexValid(index)) throw ior; Node&lt;T&gt; *temp = _head; for(int i=1; i&lt;=index; i++) { temp = temp-&gt;Next; } return temp-&gt;Value; } catch(exception&amp; e) { cout &lt;&lt; e.what() &lt;&lt; endl; } } </code></pre> <p>In my main function I have the following line:</p> <pre><code>int foo = list[5]; </code></pre> <p>Everything looks good to me, but when I compile I get the following error:</p> <pre><code>error C2440: 'initializing' : cannot convert from 'LinkedList&lt;T&gt;' to 'int' </code></pre> <p>Does anyone have any ideas how to fix this?</p> <p>Thank you!</p> <p>For those that have asked, here is how the Node class is defined:</p> <pre><code>template&lt;class T&gt; class Node { public: Node(); Node(const T value); T Value; Node&lt;T&gt; *Prev; Node&lt;T&gt; *Next; }; </code></pre> <p>And here is the declaration for my list variable:</p> <pre><code>LinkedList&lt;int&gt; *list = new LinkedList&lt;int&gt;(); </code></pre>
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    1. USsbrauen
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    1. COWhich line does the error refer to? Also, how is `list` declared?
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      1. USDrew McGowen
    2. COI would much prefer that the parameter to `operator[]` is of type `size_t`, rather than `int`.
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      1. USJohn Dibling
    3. COWhat is the definition of `list` variable? Where do you call the line `int foo = list[5];`? And why your function returns nothing if the exception was caught? Also, is `count` your alias for `cout`? :)
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      1. USpodkova
 

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