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copied!<p>joshperry's answer is very clever and elegant, but (as it is stated below the post) it doesn't check the signature of foo() properly and doesn't work with fundamental types (like int): it causes a compiler error. I will propose a technique that handles inherited members correctly and also checks the signature of the member function. Instead of going into details I will give you two exampes and hope that the code will speak for itself.</p> <h2>Example1:</h2> <p>We are checking for a member with the following signature: <code>T::const_iterator begin() const</code></p> <pre><code>template<class T> struct has_const_begin { typedef char (&Yes)[1]; typedef char (&No)[2]; template<class U> static Yes test(U const * data, typename std::enable_if<std::is_same< typename U::const_iterator, decltype(data->begin()) >::value>::type * = 0); static No test(...); static const bool value = sizeof(Yes) == sizeof(has_const_begin::test((typename std::remove_reference<T>::type*)0)); }; </code></pre> <p>Please notice that it even checks the constness of the method, and works with primitive types, as well. (I mean <code>has_const_begin<int>::value</code> is false and doesn't cause a compile-time error.) </p> <h2>Example 2</h2> <p>Now we are looking for the signature: <code>void foo(MyClass&, unsigned)</code></p> <pre><code>template<class T> struct has_foo { typedef char (&Yes)[1]; typedef char (&No)[2]; template<class U> static Yes test(U * data, MyClass* arg1 = 0, typename std::enable_if<std::is_void< decltype(data->foo(*arg1, 1u)) >::value>::type * = 0); static No test(...); static const bool value = sizeof(Yes) == sizeof(has_foo::test((typename std::remove_reference<T>::type*)0)); }; </code></pre> <p>Please notice that MyClass doesn't has to be default constructible or to satisfy any special concept. The technique works with template members, as well.</p> <p>I am eagerly waiting opinions regarding this.</p>

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