如何在C ++中调用可变参数模板构造函数？

Question

I am learning c++ template technique.

I made a constructor with variadic template augments like following.

The class is simplified very much, so it does not have meaning. However I could not call the function. Compiler tells me that it cannot call constructor directly.

How can I call it?

#include <utility>

class TemplateVariadicAugments {
public:

    template <typename FutureInnterTemplateClass, typename... Args>
    TemplateVariadicAugments(Args&&... args) : value_(std::forward<Args>(args)...) {}

    virtual ~TemplateVariadicAugments() = default;

    int value_;
};

void test_variadic_template_augments(void) {

    TemplateVariadicAugments a = TemplateVariadicAugments::template TemplateVariadicAugments<int, int>(1);

}

Answer 1

In the form you wrote it, there is no way to call the constructor. First, in C++ you cannot call a constructor by it's name, even in trivial cases:

class A
{
public:
    A() {}
};

void foo()
{
    A::A(); // Illegal.
}

Then, as you can't call the constructor directly, you can't instantiate the template explicitly, so all template arguments must be deduced. But in your case FutureInnterTemplateClass cannot be deduced, as it's not used anywhere in the constructor.

The solution is to remove redundant argument from the constructor:

template <typename... Args>
TemplateVariadicAugments(Args&&... args) : value_(std::forward<Args>(args)...) {}

Now an object may be constructed as follows:

TemplateVariadicAugments obj(1);

and Args in this case wil lbe correctly deduced to int .

But in this case it's unclear what did you want to say with variadic template argument, as if you construct an object like this:

TemplateVariadicAugments obj(1, 2);

your value_ member initialization will become equivalent ti this code:

int value_(1, 2);

which is obviously ill-formed.