Question

Related question: template-function-is-same-in-template-classes

I’m a bit disconcerted with the type of the pointer “this” (gcc 4.7.2, c++11). In principle, the type of pointer “this” of a non-const object of type C, for example, is “C * const”, and thus, the type of “*this” is “C”. But the behaviour of the “is_same” class confused me.

Test:

// this-type.cpp
#include <iostream>
#include <type_traits>

using namespace std;

class C
{
public:
   void test()
   {
       cout << boolalpha;

       cout << "'this' const?" << "              "
            << is_const<decltype(this)>::value << endl;

       cout << "'*this' const?" << "             "
            << is_const<decltype(*this)>::value << endl;

       cout << "'*this' and 'C' the same?" << "  "
            << is_same<decltype(*this), C>::value << endl;

       cout << "'this' and 'C*' the same?" << "  "
            << is_same<decltype(this), C*>::value << endl;
   }
};

int main()
{
    C c;

    c.test();
}

Output:

$ g++ --version | grep g++
g++ (Ubuntu/Linaro 4.7.2-2ubuntu1) 4.7.2
$ g++ -std=c++11 this-type.cpp
$ ./a.out
'this' const?              false
'*this' const?             false
'*this' and 'C' the same?  false
'this' and 'C*' the same?  true

However, the expected output is:

$./a.out
'this' const?              true   // C* const
'*this' const?             false  // C
'*this' and 'C' the same?  true
'this' and 'C*' the same?  false  // C* const vs. C*

What are happened here?

Answer 1

In this case, the type of this is C *

9.3.2 The this pointer [class.this]

In the body of a non-static (9.3) member function, the keyword this is a prvalue
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*.

…

No mention for X *const, etc. The this pointer is a prvalue, that’s why it cannot be assigned to, not because it const-qualified.

PS. By the way, it seems

C *p; // for any type C
is_same<declype (*p), C &>::value == true

though it may be an artefact of the implementation (compiler or is_same), because the standard says:

5.3.1 Unary operators [expr.unary.op]

1 …

… If the type of the expression is “pointer to T,” the type of the result is “T.” …