你的动作不是移动而是复制:

Your move doesn't move but copy:

A::A(A&& a) : p(a.p), s(a.s)
{
    a.s = nullptr;
    std::cout << "Move constructor\n";
}

A& A::operator=(A&& a)
{
    std::cout << "Move overload operator=\n";

    if (this != &a) {
        p = a.p;
        delete [] s;
        s = a.s;
        a.s = nullptr;
    }
    return *this;
}

现在，关于

A make_A()
{
    A a(2,"bonapart"); // Constructor
    return a;
}

由于潜在的复制省略(NRVO)，有几种情况(gcc 有标记为 -fno-elide-constructors 来控制)

There are several scenario because of potential copy elision (NRVO) (gcc has flag as -fno-elide-constructors to control that)

如果 NRVO 适用，则 a 是构造就地"；所以不会发生额外的破坏/移动；

if NRVO apply, then a is construct "in-place" so no extra destruction/move happens;

否则有一个移动构造函数和a的销毁.

else there is a move constructor and the destruction of a.

A make_A()
{
    A a(2,"bonapart"); // #2 ctor(int const char*)
    return a; // #3 move (elided with NRVO)
} // #4 destruction of a, (elided with NRVO)

int main()
{
    A a1; // #1: default ctor
    a1 = // #5: Move assignment (done after make_A)
      make_A(); // #6: destructor of temporary create by make_A

    
    a1.display();
} // #8: destructor of a1

使用 NRVO

default ctor
ctor(int const char*)
move assignment
destructor
display
destructor

没有 NRVO (-fno-elide-constructors)

default ctor
ctor(int const char*)
move ctor
destructor
move assignment
destructor
display
destructor

演示

为了

A a1,a2;
a2 = a1 = make_A();

a1 = make_A(); 使用移动赋值.a2 = (a1 = make_A()) 使用复制赋值作为移动赋值返回(正确)A&

a1 = make_A(); use move assignment. a2 = (a1 = make_A()) use copy assignment as move assignment returns (correctly) A&

4在 Move 构造函数和 Move 重载 = 操作符中，我使用了 a.s=nullptr; 这个语句总是在 Move 语义中使用 fredoverflow(user) 解释了类似现在源不再拥有它的对象"之类的东西.但我不明白.因为如果我不写这个语句仍然没有问题一切正常.请解释这一点

4 In Move constructor and Move overloaded = operator I used a.s=nullptr; This statement is always used in Move semantics fredoverflow(user) explained something like "now the source no longer owns the object it" but I am not getting it. Because if I did not write this statement still no problem everything works fine. please explain this point

你的问题是你复制而不是移动.

Your issue is that you do copy instead of move.

如果你做 s = a.s; 而不是复制

If you do s = a.s; instead of the copy

s = new char[strlen(a.s) + 1];
strcpy(s, a.s);

那么 this->s 和 as 都指向相同的数据，this 和 a> 会在它们的析构函数中释放(相同的)内存 ->双重免费错误.

then both this->s and a.s would point of same data, and both this and a would free the (same) memory in their destructor -> double free error.

a.s = nullptr; 可以解决这个问题.