在可变参数模板参数之后，C ++是否允许正常参数？

Question

According to cppreference , the following code is legal:

lock_guard( MutexTypes&... m, std::adopt_lock_t t );

However, the following code cannot be compiled with clang 3.8 (-std=c++1z):

template<typename... Args>
void f(Args&&..., bool)
{}

int main()
{
    f(1, 2, 3, true); // error! see below for details.
}

 1>main.cpp(59,2): error : no matching function for call to 'f' 1> f(1, 2, 3, true); 1> ^ 1> main.cpp(54,6) : note: candidate function not viable: requires 1 argument, but 4 were provided 1> void f(Args&&..., bool) 1> ^ 1> 1 error generated. 

Does C++ allow normal parameters after variadic parameters?

Answer 1

The function declaration in your code is valid, however deduction does not work properly for such function templates. Note that the following code is well-formed, and instantiates the specialization void f(int, int, int, bool) :

template<typename... Args>
void f(Args&&..., bool) {}

int main() {
    f<int, int, int>(1, 2, 3, true);
}

Note that in C++17, MutexTypes... are the template parameters to the class itself:

template <class... MutexTypes> class lock_guard;

so they are known and do not need to be deduced. Note that the constructor with adopt_lock_t cannot be used for C++17 class template argument deduction since the adopt_lock_t argument occurs after the parameter pack. If the committee had been prescient in C++11, they would have put the adopt_lock_t argument at the beginning rather than at the end, but alas, it is too late now.