获取this指针的地址是可以的;)
而且在类X的非const成员函数里，this的类型应该是X* const，所以可以修改this说指向的地址的值的内容，而不能修改自身；
在类X的const成员函数里，则是const X * const所以两边都不能修改  回复  更多评论
  

C++标准的确就是我文章中这样说的  回复  更多评论
  

查了下，还真的有，不过你用编译器试试就可以知道了
In the body of a nonstatic (9.3) member function, the keyword this is a non-lvalue expression whose
value is the address of the object for which the function is called. The type of this in a member function
of a class X is X*. If the member function is declared const, the type of this is const X*, if the member
function is declared volatile, the type of this is volatile X*, and if the member function is
declared const volatile, the type of this is const volatile X*.

[测试代码]
#include <iostream>
using namespace std;
class Test
{
public:
void print();
void print_const() const;
};

void Test::print()
{
cout << this << endl;
this = 1;
}

void Test::print_const() const
{
cout << this << endl;
this = 1;
};

int main()
{
return 0;
}

[编译错误]
main.cpp(13) : error C2440: '=' : cannot convert from 'int' to 'Test *const '
Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
main.cpp(19) : error C2440: '=' : cannot convert from 'int' to 'const Test *const '
Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast

VC6.0，2005都是一样，G++提示non-lvalue in assignment
正如C++标准草案中写的,the keyword this is a non-lvalue expression ,它是不能进行赋值的  回复  更多评论
  

@nono
谢谢你的指导，的确如你所说  回复  更多评论