Template call: Actual specialization not called

#include <iostream>

using namespace std;

template<typename T>
void test() {
   cout << "1";
}

template<>
void test<std::string>() {
   cout << "2";
}开发者_StackOverflow中文版

int main() {
   test<std::string()>(); //expected output 2 but actual output 1
}

Why is the output 1 and not 2?

test<std::string> (note: no parentheses at the end) would yield what you expect.

Writing it as test<std::string()> instantiates the template with the type "function taking no arguments and returning std::string"

Did you mean to invoke the function like: test<std::string>()?

In your test<std::string()>(), the template parameter is not std::string but a function type (a function taking no arguments and returning std::string).

std::string() is a typeid. A typeid is a simple declaration with missing declarator id.

In a template-argument, if there is an ambiguity between a type-id and an expression the call is resolved to a type-id. So your code outputs 1

You need to remove the parentheses () in order to get 2 as the output i.e foo<std::string>() would give you output 2.