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为了在处理数组成员时理解移动运算符，我编写了以下开发测试。

当我通过不实现赋值或移动赋值运算符而违反了 5 规则时，为什么这会起作用和/或编译？

我目前正在使用 Visual Studio 2015。

#include <cstring>
#include <iostream>
#include <string>

struct Foo
{
    Foo(const char * cstring, const std::string & string)
        :
        m_string(string)
    {
        strncpy_s(m_cstring, sizeof(m_cstring), cstring, _TRUNCATE);
    }

    Foo(const Foo & rhs)
        :
        m_string(rhs.m_string)
    {
        std::copy(rhs.m_cstring, rhs.m_cstring + sizeof(rhs.m_cstring), m_cstring);
    }

    Foo(Foo && rhs)
        :
        m_string(std::move(rhs.m_string))
    {
        std::copy(rhs.m_cstring, rhs.m_cstring + sizeof(rhs.m_cstring), m_cstring);
    }

    char        m_cstring[6];
    std::string m_string;
};

int main()
{
    // Initialize
    Foo a("Hello", "World");

    // Copy constructor
    Foo b(a);

    std::cout << &a.m_cstring << " " << a.m_cstring << " " << a.m_string << std::endl;
    std::cout << &b.m_cstring << " " << b.m_cstring << " " << b.m_string << std::endl << std::endl;

    // Move constructor
    Foo c(std::move(a));

    std::cout << &c.m_cstring << " " << c.m_cstring << " " << c.m_string << std::endl << std::endl;

    // Assignment operator
    Foo d = c;

    std::cout << &c.m_cstring << " " << c.m_cstring << " " << c.m_string << std::endl;
    std::cout << &d.m_cstring << " " << d.m_cstring << " " << d.m_string << std::endl << std::endl;

    // Move Assignment operator
    Foo e = std::move(c);

    std::cout << &e.m_cstring << " " << e.m_cstring << " " << e.m_string << std::endl << std::endl;

    return 0;
}
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