类定义中的sizeof（* this）

Question

Can we do something like this:

#include <iostream>

class Foo
{
public:
   Foo() { std::cout << sizeof(*this) << '\n'; }
};

In C Standard i see the following:

ISO/IEC 9899:2011

6.7.2.1 Structure and union specifiers

8 ... The type is incomplete until immediately after the } that terminates the list, and complete thereafter.

But in C++ Standard i can't find any analogue.

The sizeof operator shall not be applied to an expression that has incomplete type, so can we write such code or not?

Answer 1

Yes, you can write such code because the compiler has to treat it as though the class definition is complete inside class method implementations.

For example it has to treat it as though you wrote:

#include <iostream>

class Foo
{
public:
   Foo();
};

// Methods declared in the body of a class are implicitly inline
// Inline, however, probably doesn't mean what you think it means:
inline Foo::Foo() { std::cout << sizeof(*this) << '\n'; }

Answer 2

在成员函数体内部，类是完整的 - 否则，您无法访问任何其他成员函数，也无法访问任何成员变量，这将成为一个相当无价值的成员。

Answer 3

Yes you can write and also the result will also be correct.

#include<iostream>
using namespace std;
class Foo
{
public:
       int x;
   Foo() { std::cout << sizeof(*this) << '\n'; }
};
int main(){
    Foo b;
    b.x=5;
    system("pause");
    return 0;
    }