Change the value of a const variable.

[robgeek]

Hello!
Despite of the "const" in the variable declaration, why can't I change the value of "x" if did that casting to(int*)? The address of "x" and the content of "ptr" are the same.

If I remove that "const" of the declaration I can do what I want and the content of "x" and "ptr" are the same, as well as before removing "const". I don't understand!

I'm using g++ (GCC) 8.2.1 20181127.
I don't really want to do that I'm just studying and trying to understand.

#include <iostream>

int main(int argc, char const *argv[])
{
	const int x = 90;
	int* ptr = new int;
	
	ptr = (int*)&x;
	*ptr = 50;

	std::cout << "The address of x is: " << &x << "\n";
	std::cout << "The content of ptr is: " << ptr << "\n";
	std::cout << "The value of x is: " << x << "\n";

	return 0;
}

[Duthomhas]

Your question is unclear.

In C and C++, const does not necessarily mean that a value is immutable. It means that you promise not to attempt modifying the object.

Hence the reason that casting away constness is Evil^TM.

Hope this helps.

[mbozzi]

A program that modifies the value of an object declared const has undefined behavior. This is why your program's behavior makes no sense.

This snippet is fine:

int x = 90;
int const* ptr = &x;
*const_cast<int*>(ptr) = 100;

Because the object being modified (x) is not declared const.

Very strongly prefer C++-style explicit casts (e.g., const_cast over C's casts).

Hopefully it's obvious that a stray const_cast<int*>(ptr) is more likely to draw careful scrutiny than a C-style cast, which does not make it clear that cv-qualifiers are being removed.

Evil^TM or not, casting away const can be substantially useful (for example, to reduce code duplication or to interface with const-incorrect code), but it is best avoided.