Programming

See code below. It appears that i and j inhabit the same address yet hold different values. Can anyone shed light on this?

Code:

int main() {
  const int i= 3;
  int* j = const_cast<int*>(&i);
  *j = 5;
  cout << j << endl << &i << endl;
  cout << *j << endl << i;
}

Interesting. What compiler are you using?

I understand that if you cast away the constness of an object that has been explicitly declared as const, and attempt to modify it, the results are undefined.

I'm using GNU GCC compiler.

If an operation is specified as having an undefined result in the language standard, does the compiler documentation tell how they implemented it?

"undefined result" means "don't do that".

what does your code actually print?

Typically, in terms of conformance to a standard, undefined means the value or behavior may vary among implementations that conform to a standard. An application should not rely on the existence or validity of the value or behavior. Any application that relies on any particular value or behavior cannot be assured to be portable across conforming implementations.

Here is the example

Code:

#include <iostream>

using namespace std;

int main() {
  const int i= 3;
  int* j = const_cast<int*>(&i);
  cout << j << endl << &i << endl;
  *j = 5;
  cout << *j << endl << i << endl;
}

whose output is:

Code:

0x81ffb0
0x81ffb0
5
3

In general, const_cast is safe only if you're casting a variable that was originally non-const.