C++ 模板定义 function 的参数个数

Question

Lets say i have a class looking like this

template<int n>
class A{
   array<size_t, n> sizes;
   //...
public:
   template <int k>
   A<k> reshape(array<size_t, k> new_sizes){
      return A<k>(new_sizes):
   }
};

it works but the parameter new_sizes is syntatically suboptimal, since i have to call it like that:

foo.reshape(array<size_t, 3>{1,2,3});

This does not work:

foo.reshape({1,2,3});

Is there a way to either define a initializer_list with compile time size (so i could use it instead of the array) OR (even better) a way to define the size of a variadic parameter, so i could write something like

foo.reshape(1,2,3);

Answer 1

OR (even better) a way to define the size of a variadic parameter, so i could write something like foo.reshape(1,2,3);

You could take the sizeof... a parameter pack:

template <size_t N>
class A {
    std::array<size_t, N> sizes;

public:
    A() = default;

    template <class... Args>
    A(Args&&... ss) : sizes{static_cast<size_t>(ss)...} {}

    template <class... Args>
    A<sizeof...(Args)> reshape(Args&&... new_sizes) {
//    ^^^^^^^^^^^^^^^
        return A<sizeof...(Args)>(static_cast<size_t>(new_sizes)...);
    }
};

// deduction guide:
template <class... Args>
A(Args&&...) -> A<sizeof...(Args)>;

Demo

Answer 2

{1, 2, 3} has no type but can be deduced as initializer_list<T> or T[N]

So to keep your syntax, it would be:

template <std::size_t K>
A<K> reshape(/*const*/ size_t (&new_sizes)[K])
{
    return A<K>(new_sizes):
}

so i could write something like foo.reshape(1,2,3);

For that variadic template:

template <typename... Ts>
A<sizeof...Ts> reshape(Ts... elems)
{
    return reshape({elems...}); // using above method
}