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    <p>I'll forget that <code>array[0] = A item1;</code> isn't valid C++ and just assume that you're assigning an object of type <code>A</code> to <code>array[0]</code> and an object of type <code>B</code> to <code>array[1]</code>. Okay, so you have two problems.</p> <p>The first is known as object slicing. When you copy an object of type <code>B</code> to an object of type <code>A</code>, you only copy the <code>A</code> part of that object. So what you have in <code>array[1]</code> is not a <code>B</code> at all, it's just an <code>A</code>. If you want polymorphism (which you do), then you need to use either pointers or references which provide polymorphic behaviour. That means make your array an <code>A* array[2];</code> and do <code>array[0] = &amp;item1; array[1] = &amp;item2;</code>.</p> <p>Now, when you call a function on a pointer to <code>A</code> that points to a <code>B</code> it will still only call <code>A</code>s <code>foo</code> member function. Why? Because by default, the function will be looked up on the <em>static type</em> of the object. That static type is <code>A</code>. If you want to tell the compiler to look up your function on the <em>dynamic type</em> of your object - the <em>true</em> type of your object, which is <code>B</code> - you need to make that member function virtual. So in <code>A</code>, do:</p> <pre><code>virtual int foo(int c) { c = c+1; return c; }; </code></pre> <p>Now when your compiler see that you're calling <code>foo</code> on an <code>A*</code>, it'll see that it's virtual and say "Oh okay, I should look up this function dynamically" and it'll find <code>B</code>'s implementation of <code>foo</code>.</p>
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    1. USJoseph Mansfield
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    1. COThanks a lot! I knew that was inappropriate syntax but thanks for calling it out. I applied your suggestions and the code ran just like I wanted to.
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      1. USRyan Tuck
 

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