为什么构造函数调用取决于默认析构函数的存在？

Question

In the following code, I get two constructor calls for Test u = "u"; . However, if I comment out the destructor, then I only get one constructor call. Why is that?

#include <iostream>

template<class T>
auto operator<<(std::ostream& os, const T& t) -> decltype(t.print(os), os) 
{ 
    t.print(os); 
    return os; 
} 

class Test 
{
public:
    template<typename T>
    Test(T&& t)
    {
        std::cout << "Test " << t << '\n';
    }
    ~Test() = default; // if commented out removes one construction
    void print(std::ostream& os) const
    {
        os << "[with T = Test]";
    }
};

int main()
{
    Test u = "u"; // two constructors (second, a temporary, with T = Test)
    Test t("t"); // one constructor
}

Answer 1

The user-declared destructor, even if it is defaulted, means no move constructor gets generated.

12.8 Copying and moving class objects [class.copy]

9 If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if

[...]

(9.4) -- X does not have a user-declared destructor.

If a move constructor is generated, then it is a better candidate for moves than your template constructor. It doesn't save a constructor call, it just means a different constructor gets called, one that doesn't print anything.