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copied!<p><a href="https://stackoverflow.com/questions/3215756/operator-in-subclass/3215856#3215856">Martin told you what went wrong</a>, here's what you should do instead: </p> <pre><code>Bar bar; Foo& foo = bar; // this works now </code></pre> <p>Now <code>foo</code> is a <strong><em>reference</em></strong> to a an object, and you can have base class references (and pointers, BTW) refer to objects of derived classes. </p> <p>However, this</p> <pre><code>bar = foo; </code></pre> <p>will never work (at least not implicitly) and it shouldn't. What you're trying to do here is to assign a base class object (or reference) to a derived class.<br> However, while a derived class can always stand in for a base class object, the opposite is not true. (You can't just mindlessly use any <strong><em>vehicle</em></strong> when what you need is a <strong><em>boat</em></strong>, because not all vehicles are boats, the vehicle you're using might be a <strong><em>car</em></strong> and you'd drown.) </p> <p>This is not to say that you cannot make it work. if you know how to create a derived class object from a base class object, you can write a function to do so. It's just that the compiler doesn't know how to make a <em>car</em> from a <em>vehicle</em>. </p> <p>As <a href="https://stackoverflow.com/questions/3215756/operator-in-subclass/3216152#3216152">solipist has shown</a>, such a conversion function can be a constructor. However, I would make conversion constructors explicit: </p> <pre><code>class Bar: public Foo { public: explicit Bar(const Foo&); // } </code></pre> <p>The <code>explicit</code> keyword makes sure that the compiler will never attempt to call this constructor unless you say so. If you remove it, then this code would compile: </p> <pre><code>//void f(Foo); // you meant to call this, but forgot to include its header void f(Bar); // you didn't even know this was included Foo foo; f(foo); // won't compile with the above Bar class </code></pre> <p>and the compiler would silently generate this code: <code>f(Bar(foo));</code><br> While this seems handy at first, I think I have sooner or later been bitten by every single implicit conversions I allowed to creep into my code and had to remove them later. So many years ago I swore to never allow them anymore.<br> Note that even with the <code>explicit</code> conversion constructor you can still call <code>f(Bar)</code> with a <code>Foo</code> object - you just have to say so explicitly: </p> <pre><code>f(Bar(foo)); // compiles fine even with the explicit conversion constructor </code></pre>

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