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Following Template class type-specific functions, how can I customize my template code to not compile for certain types? If the question is not clear, take a look at this example.

///** template class ***/ template<typename T> class testClass{ testClass(); T parameter; } template<typename T> void testClass<T>::print(){cout<<parameter.value<<endl;}

The above class is supposed to work for the following types:

 //** types file **/ class paramtype1{ int value; } class paramtype2{ int value; } class paramtype3{ }

As you see, paramtype3 doesn't have value, so I get a compile error saying that value is not defined. I know that if I want to specialize a template class function for a certain type (s), I need to do:

template<> void testClass<paramtype1>::print(){cout<<parameter.value<<endl;}

But, is there any way to do the other way around, only excluding some certain types?

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  • You should edit your question to include what types specifically you want to include/exclude, and why. If you want to enable a specialization only if a type has a particular member, you'd use different techniques. Commented Jul 25, 2019 at 12:11
  • Also, you got a compiler error when you tried to use the template with paramtype3... doesn't that mean it already didn't compile for that type? Commented Jul 25, 2019 at 12:13

2 Answers 2

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If you want enable/disable the full class/struct, you can use SFINAE and partial specialization.

The following is a C++17 example

template <typename T, typename = void> struct testClass; template <typename T> struct testClass<T, std::void_t<decltype(T::value)>> { testClass() { }; T parameter; void print() { std::cout << parameter.value << std::endl; } };

If you only want enable/disable the print() function, you have to templatize it; by example

template <typename U = T> std::void_t<decltype(U::value)> print() { std::cout << parameter.value << std::endl; }

or also

template <typename U = T> std::void_t<decltype(U::value), std::enable_if_t<std::is_same_v<U, T>>> print() { std::cout << parameter.value << std::endl; }

if you want to be sure that nobody can "hijack" the method explicating the template type calling

testClass<paramtype3>{}.print<paramtype1>():
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What I would personally do to exclude permissions to use certain types is:

template <class T, class... Ts> struct is_any : std::disjunction<std::is_same<T, Ts>...> {}; // https://stackoverflow.com/questions/17032310/how-to-make-a-variadic-is-same template <typename T> void do_something() { static_assert(!is_any<T, int, bool>::value, "do_something<T> cannot be used with T as int or bool"); // code here }

Allows you to add a custom assertion message aswel, making it easy to realise what's wrong.

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