我的 Copy and Move 构造函数和赋值运算符出现问题

问：

我想用 C++ 构建自己的完整类。我是这样开始的：Vector

#include <iostream>
#include <initializer_list>

#define Print(x)(std::cout<< x << std::endl)

// Vector Class // 
template <typename T>
class Vector
{
    // Attributes
    int length = 0;
    T* array;

    public:
   
    // Default constructor
    Vector()
    : length(0), array(nullptr)
    { }

    // Copy constructor 
    template<typename U>
    Vector(const Vector<U>& other)
        : Vector(other.len())
    {
        Print("Use Copy Constructor");
        // Coppying other data to new array
        array = new T[other.len()];
        for (auto i=0; i < other.len(); i++)
            array[i] = other[i];
    }

    // Move constructor
    Vector(Vector<T>&& other)
        : length(other.len()), array(other.array)
    {
        Print("Use Move Constructor");
        // Deleting other array
        other.length = 0;
        other.array = nullptr;
    }

    // Constructor (does not allow uniform initialisation)
    Vector(int length)
    : length(length), array(new T[length])
    { }

    // Constructor (with initialiser list and delegate constructor)
    Vector(std::initializer_list<T> list)
    : Vector((int)list.size())
    {
        std::uninitialized_copy(list.begin(), list.end(), array);
    }

    // Destructor
    ~Vector()
    {
        length = 0;
        delete[] array;
        array = nullptr;
    }

    // Function Len that returns the length 
    int len() const
    {
        return length;
    }
    // Operator[](int i) and its const overload
    auto& operator[](int i)
    {
        return array[i];
    }
    
    auto& operator[](int i) const
    {
        return array[i];
    }

    // Copy assignment operator 
    template<typename U>
    Vector& operator=(const Vector<U>& other)
    {
        Print("Use Copy Operator");
        if (this != (Vector*)& other) {
            /*
            This works the same way but does not solve the problem: 
    
            Vector<typename std::common_type<T,U>::type> temp(other);
            temp.swap(*this); 
            */ 

            delete[] array;
            length = other.len();
            array = new T[other.len()];
            for (auto i = 0; i < other.len(); i++)
                array[i] = other[i];
        }
        return *this;
    }

    // Move assignment opertator
    Vector& operator=(Vector<T>&& other)
    {
        Print("Use Move Operator");
        if (this != (Vector*)&other){
            /*
            This works the same way but does not solve the problem: 

            Vector<T> temp(std::move(other)); // moves the array
            temp.swap(*this);
            */ 

            delete[] array;
            length = other.len();
            array   = other.array;
            other.len() = 0;
            other.array   = nullptr;
        }
        return *this;
    }
};

但是，如果我现在尝试像这样使用复制分配运算符：

Vector<double> double_vector = {3.4677, 3.212, 2.2, 6.3};
Vector<double> a = double_vector;

我收到以下错误消息：

错误：使用已删除的函数“constexpr Vector<double>：：Vector（const Vector<double>&）”

我认为问题出在复制构造函数和复制赋值运算符上，但遗憾的是我找不到解决方案。我觉得奇怪的是，如果我注释掉移动构造函数和移动赋值运算符，代码似乎确实有效。这让我认为编译器很难知道要使用哪个构造函数。但我也可能对此大错特错。

希望我提供了足够的信息来回答/推动正确的方向。