How can SFINAE be used effectively with member functions of class templates?

SFINAE for Member Functions of Class Templates

In the provided code, SFINAE is applied to the member functions of a class template. However, this approach encounters an error because SFINAE is only applicable to deduced template arguments, specifically for function templates.

In this case, the class Foo is a template, but both member functions bar() are unconditionally instantiated regardless of the template parameter T. This leads to the compilation error indicating that the two overloads cannot be both valid.

To resolve this issue, SFINAE can be used correctly by deducing the template arguments for the member functions. The following changes can be made:

<code class="cpp">#include <type_traits>

struct A{};
struct B{};

template <typename T>
struct Foo
{
    template <typename U = T>   // Deduce the template argument
    typename std::enable_if<std::is_same<U, A>::value>::type
    bar() {}

    template <typename U = T>   // Deduce the template argument
    typename std::enable_if<std::is_same<U, B>::value>::type
    bar() {}
};</code>

By deducing the template arguments for the member functions, SFINAE can now determine which overload of bar() to instantiate based on the actual template argument provided for Foo. This approach ensures that the code will compile successfully and behave as intended.

The above is the detailed content of How can SFINAE be used effectively with member functions of class templates?. For more information, please follow other related articles on the PHP Chinese website!