const_cast's (again)

Some of the previous threads about const_casts have left me wondering about something. So I wrote the following little test program, the output of which surprised me a little.

Code:

---

#include <iostream>

void increment( int& n )
{
 ++n;
}

int main()
{
 int i = 1;
 std::cout << i << " - ";
 increment( i );
 std::cout << i << "\n";

 const int j = 1;
 std::cout << j << " - ";
 // The following does not compile
 // int l = const_cast<int>( j );
 int * pj = const_cast<int*>( &j );
 increment( *pj );
 std::cout << j << "\n\n";
 return 0;
}

---

Output:
1 - 2
1 - 1

Does const_cast create a copy of the object? Or is this what "undefined behavior" stands for?

Let me know if other compilers (I'm using g++) create different results.

Hello ,

The out put u can't say wht exactly it will come tht behavoir is called Undefined behavior

I'll try answering this with entire words and complete sentences.

Quote:

---

Originally posted by treuss
Does const_cast create a copy of the object? Or is this what "undefined behavior" stands for?

Let me know if other compilers (I'm using g++) create different results.

---

This is a completely invalid thing to do. A const POD item will not likely exist in normal variable space. At best, you are modifying non-modifyable memory. Even worse, a compiler could do away with this definition completely so that it takes up 0 memory.

Of course, it's compiler dependent. The behavior is indeed undefined.

Jeff

Hi,
To be precise the problem is in the statement

Code:

---

                increment( *pj );

---

You are passing some temporary variable to the increment function.

And n (in increment) would now be an alias to this variable and not *pj.

Thus this seems to me as a defined behaviour as far as value of *pj is concerned, since there is no reason that it should change.

Regards,