Type trait dependent specialization of template function

Question

I have a simple templated function, defined in a library I am using

template<class T>
T create(std::vector<char>& data)
{ 
    T newValue{}; 
    /* Do something with data */
    return newValue;
}

and I want to specialize this function in case T implements a specific interface

template<class T>
std::enable_if_t<std::is_base_of<Interface, T>::value, T> create( std::vector<char>& data)
{ 
    T newValue{};
    newValue.InterfaceFunction(data);
    return newValue;
}

but I cannot make this work, the function I have specialized is not used. How can I achieve making a specialization of an already defined template function?

I think this is the reason why it is not working, you are not technically specializing the first function if you still use the T class... the compiler sees the second one as a totally different function — Oneiros, Jul 03 '18 at 07:22
Sorry, the example was bad, I have rewritten it. Is it still not technically possible? — thorsan, Jul 03 '18 at 07:24
Does [this thread](https://stackoverflow.com/questions/31500426/why-does-enable-if-t-in-template-arguments-complains-about-redefinitions) help? — Oneiros, Jul 03 '18 at 07:24
@Oneiros I have tried that way, but then it becomes ambigous. — thorsan, Jul 03 '18 at 07:30

songyuanyao · Accepted Answer · 2018-07-03T08:05:51.690

1

This is not template specialization but templates overloading, and function templates can't be partial specialized. The problem is that when you specify a type deriving from Interface both the function templates are exact match, which leads to ambiguity.

You can apply SFINAE.

template<class T>
std::enable_if_t<!std::is_base_of<Interface, T>::value, T> create(std::vector<char>& data)
{ 
    T newValue{}; 
    /* Do something with data */
    return newValue;
}

template<class T>
std::enable_if_t<std::is_base_of<Interface, T>::value, T> create( std::vector<char>& data)
{ 
    T newValue{};
    newValue.InterfaceFunction(data);
    return newValue;
}

LIVE

edited Jul 03 '18 at 08:05

answered Jul 03 '18 at 07:29

songyuanyao

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Thank you for your answer. I know it can be done this way, IF i was allowed to change the original definition, but I am not. – thorsan Jul 03 '18 at 07:31
@thorsan I added some explanation about why the original definition doesn't work. – songyuanyao Jul 03 '18 at 07:37
@thorsan: If possible, you may still create `my_create` which calls unmodifiable `create` for non interface, else the specialized function. – Jarod42 Jul 03 '18 at 10:20
@Jarod42, yes, that is what I ended up doing, creating a wrapper function with SFINAE, as in the above answer, selecting the right one. – thorsan Jul 04 '18 at 06:11

Type trait dependent specialization of template function

1 Answers1