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    <p>Sorry, but the design seems to be flawed in several aspects. First,</p> <pre><code>typedef test_output&amp; (*test_function)(test_input&amp;); </code></pre> <p>calls for trouble. Don't return references to objects generated inside a function. The referenced object is destroyed, when the called function ends, so you will store in <code>result</code> the address of an object that does not exist anymore (dangling pointer). If you need polymorphic results, return them safely as <code>shared_ptr&lt;test_output&gt;</code>, which should also be the type of <code>result</code>.</p> <p>Concerning your list of questions (sorry for long answer):</p> <blockquote> <p>to bind the equivalent of operator==(*result, expected) to the equality operator for the dynamic types of these things;</p> </blockquote> <p>Yes, as you did. The equality test should be overridden for every class derived from <code>test_output</code>.</p> <blockquote> <p>to be able to indicate to the compiler that those types are derived from test_output;</p> </blockquote> <p>For comparing polymorphic objects for the left operand with polymorphic objects on the right hand side, welcome to the field of <em>multi-methods</em>. The problem boils down to: "When do you consider two objects as equal?" Should they have exactly the same type or can one belong to a subset of the other? Normally the <em>Liskov substitution principle</em> in object orientation means: A derived object can be substituted for a base object, since inheritance guarantees it has all base properties. To determine if <code>expected</code> to be of (at least) <code>to_derived</code> you need a <em>downcast</em>: </p> <pre><code>bool to_derived::operator==(const test_output&amp; expected) { if (to_derived* e = dynamic_cast&lt;to_derived*&gt;(&amp;expected)) { // compare all data fields, then return true; } return false; } </code></pre> <p>If you want to guarantee <code>expected</code> to be of <strong>exactly</strong> the same type, you should use <a href="http://en.wikipedia.org/wiki/Run-time_type_information" rel="nofollow">RTTI</a> first:</p> <pre><code> if (typeid(*this) != typeid(expected)) return false; </code></pre> <p>Derived objects are no consired of the same type. Maybe your compiler needs a switch to enable that.</p> <blockquote> <p>to be able to reassign result.</p> </blockquote> <p>Use a <code>shared_ptr&lt;test_output&gt;</code>, because it manages memory for dynamic objects.</p> <blockquote> <p>is it valid to use those function pointers as arguments to test_object? Their arguments can be dynamically typed but I don't know whether that means the function type itself has an automatic conversion to the function type taking and returning the respective base types. If not, how would I indicate a function of this dynamically-typed signature?</p> </blockquote> <p>The function pointers does not need any conversion. But your test framework seems to be incomplete, since it may work only for standalone functions. What about methods? I would suggest a templated solution. </p> <p>I would seriously consider looking for an <a href="http://en.wikipedia.org/wiki/List_of_unit_testing_frameworks#C.2B.2B" rel="nofollow">existing unit testing framework</a>.</p>
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    1. USRené Richter
    1. USRené Richter
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    1. COYou say that `operator==(*result, expected)` *does* call `operator==(to_derived&, to_derived&)` and not `operator==(test_output&, test_output&)`? So if I have `to_derived: public test_output {int x;}` with the natural equality, and if `expected` has `x = 0` and `some_function_one` returns something with `x = 1`, does my code print the diagnostic or not? It seems to me it would not: the type of `*result` is determined at compilation to be `test_output` and so `operator==` is bound to `operator==(test_output&, test_output&)` whose obvious definition is always `true` (and ignores `x` anyway).
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      1. USRyan Reich
    2. CO@Ryan : No, `(*result).operator==(expected)` is called because `test_output::operator==(const test_output& )` was declared as *member* function. However, it's a pure virtual (abstract) method. So you *have to* implement that operator for derived classes, otherwise you couldn't instantiate `to_derived` objects. A meaningful `to_derived::operator==` function body would then compare `this->x == e->x`. Since the base class `operator==` is declared *virtual*, the type of the left operand `*result` is determined at *runtime* (late binding). That's the basic idea of dynamic polymorphism.
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    3. COWhat do you know, I've been fooled. I thought that when you dereferenced a pointer to a base type it stopped being polymorphic, but a few quick tests show that what you say is true: it still gets the right virtual method. So basically, the `*` operator returns a reference bound to the pointed-to object.
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