将可变数量的参数传递给函数

Question

我知道如何使用可变模板和省略号来接受可变数量的参数，但是你如何传递可变数量的参数函数中的参数？

以下面的代码为例：

#include <iostream>

struct A {
   A(int a, int b) : x(a), y(b) {}
   int x, y;
};

struct B {
    B(int a, int b, int c) : x(a), y(b), z(c) {}
    int x, y, z;
};

template<typename T, typename... TArgs>
T* createElement(TArgs&&... MArgs) {
    T* element = new T(std::forward<TArgs>(MArgs)...);
    return element;
}

int main() {

    int Aargs[] = { 1, 2 };
    int Bargs[] = { 1, 2, 3 };

    A* a = createElement<A>(Aargs); //ERROR
    B* b = createElement<B>(Bargs); //ERROR

    std::cout << "a.x: " << a->x << "\na.y: " << a->y << "\n" << std::endl;
    std::cout << "b.x: " << b->x << "\nb.y: " << b->y << "\nb.z: " << b->z << "\n" << std::endl;

    delete a;
    delete b;
}

有没有办法扩展数组，使它们的每个值都像传递给函数的参数（类似于参数包扩展）？

或者，如果不行，还有其他方法可以实现吗？

Answer 1

您可以使用 std::index_sequence

扩展数组

#include <iostream>
#include <utility>

struct A {
   A(int a, int b) : x(a), y(b) {}
   int x, y;
};

struct B {
    B(int a, int b, int c) : x(a), y(b), z(c) {}
    int x, y, z;
};

template<typename T, typename... TArgs>
T* createElement(TArgs&&... MArgs) {
    T* element = new T(std::forward<TArgs>(MArgs)...);
    return element;
}

template<typename T, typename U, size_t...  I>
T* createElementFromArrayHelper(std::index_sequence<I...>, U* a){
    return createElement<T>(a[I]...);
}

template<typename T, typename U, size_t N>
T* createElementFromArray(U (&a)[N]){
    return createElementFromArrayHelper<T>(std::make_index_sequence<N>{}, a);
}

int main() {

    int Aargs[] = { 1, 2 };
    int Bargs[] = { 1, 2, 3 };

    A* a = createElementFromArray<A>(Aargs); 
    B* b = createElementFromArray<B>(Bargs);

    std::cout << "a.x: " << a->x << "\na.y: " << a->y << "\n" << std::endl;
    std::cout << "b.x: " << b->x << "\nb.y: " << b->y << "\nb.z: " << b->z << "\n" << std::endl;

    delete a;
    delete b;

}

将可变数量的参数传递给函数

Pass a variable number of arguments into a function

c++

arrays

variadic-functions

variadic-templates

parameter-pack