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2012-05-08T08:33:32.803
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Apart from what @dag has mentioned, a function definition without parameter name finds its usage in template specialization. I know you have mentioned that there are no templates in the example you have posted, but for sake of completeness i'd like to post this use case: Suppose you have defined an <code>erase</code> function for <code>std</code> containers. But you want different <code>erase_helper</code> to do the actual work based on the container type. A common and acceptable practice is to use <code>traits</code>. <pre><code>// Tags for containers struct vector_tag {}; struct map_tag {}; // Trait class template <typename C> struct container_traits; // Specialization of trait class template <typename T, typename A> struct container_traits<std::vector<T, A>> { typedef vector_tag category; }; template <typename K, typename V, typename C, typename A> struct container_traits<std::map<K, V, C, A>> { typedef map_tag category; }; // Helper function template <typename Container, typename X> void erase_helper(Container& c, const X& x, vector_tag) // <-- no param name { // Erase element from vector } template <typename Container, typename X> void erase_helper(Container& c, const X& x, map_tag) // <-- no param name { // Erase element from map } // Function interface template <typename Container, typename X> void erase(Container& c, const X& x) { erase_helper(c, x, typename container_traits<Container>::category()); } </code></pre> You can see here, <code>erase_helper</code> has third parameter without a name. Type of parameter tells compiler to pick correct function at template instantiation phase.
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1. COTo be fair, this is still polymorphic usage. You're just doing your polymorphism at compile time instead of run time.
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2. CO@Managu, yes it is. But original question was not asking for all use cases except polymorphic usage :-).
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