CODEWITHSUNDEEP
c++templates
Dean
Dean

Reputation: 6958

Using template<> in argument

I'm used to the template syntax for a fully specialized template:

template<typename T>
struct S {};

template<>
struct S<int> {}; // Fully specialized

But I have no idea that I could use it as an argument:

void fn(std::function<> lambda){
}

int main() {
  fn([](){ std::cout << "Hello"; });
}

Does the above mean "give me a fully specialized std::function as a parameter" ? Why hasn't it the template <> std::function syntax?

Upvotes: 1

Views: 63

Answers (1)

Walter
Walter

Reputation: 45444

It appears (see this documentation of std::function) that the C++ standard does not provide for a fully specialized version of std::function<signature> , i.e. a default such as signature=void(). Thus, if your code compiled, then the C++ standard library used was not fully standard compliant. You should file a bug report.

Upvotes: 1

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