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POCan templates be completely avoided in C++?
 

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  1. POCan templates be completely avoided in C++?
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    <p>My question, which is in the last paragraph, needs (in my opinion) some explanatory setup. Basically, I'm wondering if you can avoid using templates if instead you make all your would-be template classes instead inherit from a base class that declares virtual methods you will be using, among them a function for memory allocation that, when implemented, will return a pointer to the derived (not base) type.</p> <h2>BEGIN setup</h2> <p>C++ does not seem to have the notion of a "universal base class" from which everything is implicitly derived; I imagine that class would be defined like this:</p> <pre><code>class universal_base { }; </code></pre> <p>Of course, now that I have defined it, I can make all my classes derive from it. Then, because of polymorphism, any references or pointers to <code>universal_base</code> that I pass around will be basically the same as template parameters:</p> <pre><code>template &lt;typename T&gt; class C { T &amp;x; int f(T &amp;y); C(T &amp;z): x(z + 1) {} }; class C { universal_base &amp;x; int f(universal_base &amp;y); C(universal_base &amp;z): x(z + 1) {} }; </code></pre> <p>The difference is that in the first construction, the expression <code>z + 1</code> can't be guaranteed to be valid; you just have to tell users that <code>T</code> must overload <code>operator+</code>. In the second construction, I could add a virtual such operator to <code>universal_base</code>:</p> <pre><code>// in universal_base public: virtual universal_base&amp; operator+(const universal_base &amp;x) = 0; </code></pre> <p>and use <code>typeid</code> and <code>dynamic_cast</code> in the implementations to get the argument right. This way, it is impossible to write ill-formed code, because the compiler will complain if you don't implement <code>operator+</code>.</p> <p>Of course, this way it is not possible to declare members of non-reference type:</p> <pre><code>class C: public universal_base { universal_base x; // Error: universal_base is a virtual type }; </code></pre> <p>However, this can be gotten around through careful use of initialization. In fact, if I wanted to create a template for the above,</p> <pre><code>template &lt;typename T&gt; class C: public universal_base { T x; }; </code></pre> <p>I would almost certainly be giving it objects of type <code>T</code> at some point. In that case, there is no reason that I could not do the following:</p> <pre><code>class universal_base { public: virtual universal_base&amp; clone() = 0; }; class C: public universal_base { universal_base &amp;x; C(universal_base &amp;y) : x(y.clone()) {} } </code></pre> <p>Effectively, I create a variable of a type that is determined at runtime. This of course requires that every object of type <code>C</code> be appropriately initialized, but I do not think this is a huge sacrifice.</p> <p>This is not academic, since it has the following use: if I am writing a module that is intended to be linked into other programs and handle their data in some generic way, I cannot possibly know the types that will be used. Templates are not helpful in this situation, but the technique above works fine.</p> <h2>END setup</h2> <p>So, my question: does this completely replace templates, modulo the thing about initialization? Is it inefficient or dangerous somehow?</p>
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    1. USRyan Reich
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    1. COSounds like a horrible idea. C++ isn't Java.
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    2. CO`C(T &z): x(z + 1) {}` - you can't bind `x` to that temporary :-/. `virtual universal_base& operator+(const universal_base &x) = 0;` - `+` should return by value.
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    3. CO@Mike: I don't know Java yet. Could you explain what makes this work in Java but not in C++?
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