獲取this指針的地址是可以的;)
而且在類X的非const成員函數里，this的類型應該是X* const，所以可以修改this說指向的地址的值的內容，而不能修改自身；
在類X的const成員函數里，則是const X * const所以兩邊都不能修改  回復  更多評論
  

C++標準的確就是我文章中這樣說的  回復  更多評論
  

查了下，還真的有，不過你用編譯器試試就可以知道了
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*.

[測試代碼]
#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;
}

[編譯錯誤]
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都是一樣，G++提示non-lvalue in assignment
正如C++標準草案中寫的,the keyword this is a non-lvalue expression ,它是不能進行賦值的  回復  更多評論
  

@nono
謝謝你的指導，的確如你所說  回復  更多評論