为什么删除复制构造函数会影响用户定义的默认构造函数?

学习C++时,对于构造函数几乎没有问题。

请考虑以下代码:

#include<stdio.h>
#include<iostream>

// Case 1

class CFoo
{
public:
    CFoo()  { printf("CFoo constructor: user-defined default\n"); }
    ~CFoo() { printf("CFoo destructor\n"); }
};

void testFoo()
{
    CFoo foo0;              // A way to use default constructor
    CFoo foo1 = CFoo();     // Another way to use default constructor
    CFoo foo2 = CFoo(foo1); // Using implicit copy constructor

    // Output:
    //     CFoo constructor: user-defined default
    //     CFoo constructor: user-defined default
    //     CFoo destructor
    //     CFoo destructor
    //     CFoo destructor
    //     CFoo destructor
}

// Case 2

class CBar
{
public:
    CBar() { printf("CBar constructor: user-defined default\n"); }
    CBar(CBar & other) = delete;
    ~CBar() { printf("CBar destructor\n"); }
};

void testBar()
{
    CBar bar0;
    // line 44: error C2280: 'CBar::CBar(CBar &)': attempting to reference a deleted function
    // line 34: note: see declaration of 'CBar::CBar'
    // line 34: note: 'CBar::CBar(CBar &)' : function was explicitly deleted
    // CBar bar1 = CBar(); // Why this has anything to do with the copy constructor?

    // Output:
    //     CBar constructor: user-defined default
    //     CBar destructor
}

// Case 3

class CBaz
{
public:
    CBaz()             { printf("CBaz constructor: user-defined default\n"); }
    CBaz(CBaz & other) { printf("CBaz constructor: user-defined copy\n"); }
    ~CBaz()            { printf("CBaz destructor\n"); }
};

void testBaz()
{
    CBaz baz0;
    CBaz baz1 = CBaz();
    CBaz baz2 = CBaz(baz1);

    // Output:
    //     CBaz constructor: user-defined default
    //     CBaz constructor: user-defined default
    //     CBaz constructor: user-defined copy
    //     CBaz destructor
    //     CBaz destructor
    //     CBaz destructor
}

// main

void main() {
    testFoo();
    testBar();
    testBaz();

    std::cin.get();
}

问题:

1. 为什么我无法创建的实例 CBar CBar bar1 = CBar(); 就像我做的那样 CFoo ?

2. testFoo 调用4个析构函数。其中3个用于foo0、foo1和foo2。第四名来自哪里? testBaz 具有相同的结构,但仅调用3个析构函数。唯一的区别是 和 CBaz 是吗 具有用户定义的副本构造函数。