Why move semantics has the same behavior as shallow copy in dynamic mem allocation?

In the classes dealing with dynamic mem allocation, shallow copy basically causes the program to delete a resource twice. In move operations, the original pointer no longer points to the resource, So But why the same behavior occurs in move semantics? e.g:

#include <utility>
#include <cstring>
using namespace std;
class MyString
{
    char* cstr;
 public:
    MyString(const char* arg)
    : cstr(new char[strlen(arg)+1])
    {
        strcpy(cstr, arg);
    }
    MyString(MyString&&) = default;
    MyString& operator=(MyString&&) = default;
    MyString(const MyString&) = delete;
    MyString& operator=(const MyString&) = delete;
    ~MyString()
    {
        delete[] cstr;
    }
};

int main()
{
    MyString S1{"aaa"};
    MyString S2 = move(S1); // error

}

I've tried with 3 different compilers and i got the same results.