mechanism for "undefined-ness" of modifying the value of a const

Question

I have read, it is undefined in some C standards (perhaps 99?) what happens when a const is modified. But a student presented me with some code, which I modified.

I cannot see anything special about the address of the constant variable a. I verified that &a and b are the same, so the compiler is not subtly pointing to some other location. Yet when I assign *b, the const value does not change.

I am not running with optimization. When I compile with the -g flag to debug and step into the code, I get the results I expect (the memory location of the variable a changes). Yet the code presented below does not reflect the updated value of a.

Is this that temps are now being placed in registers even in debug mode, with no optimization?

#include <iostream>

using namespace std;

int main(){
    const int a = 15;
    cout << a << '\n';
    int * b= (int*)&a;
    cout << &a << "\n";
    cout << b << "\n";
    *b = 20;
    cout << *b << '\n';
    cout << a << '\n';

    int x = a;
    cout << x << '\n';
    x = *b;
    cout << x << '\n';
    return 1;
}

So you acknowledge that it's undefined behavior, but then you ask why it doesn't behave as you might expect (whatever that is)? By definition, "undefined behavior" means the behavior doesn't have to make sense. — Michael Burr, Oct 10 '14 at 04:40
I would expect that if I write to a memory location, and if a variable represents that memory location, then I would have successfully created an error. If the code were optimized and the value were put elsewhere (like a register) then I would understand the mechanism. I guess asking for the address of the variable is not truly honored since it is const. — Dov, Oct 10 '14 at 13:07
We see a lot of undefined behavior question on SO, I think it is unfortunate how the community reacts to them sometimes. This is not a typical question, sometimes peaking under the hood can be informative like the [Deep C presentation does](http://www.slideshare.net/olvemaudal/deep-c). — Shafik Yaghmour, Oct 10 '14 at 13:24

Shafik Yaghmour · Accepted Answer · 2014-10-10T13:36:11.657

This is also undefined behavior in C++, we can see this by going to the draft C++ standard section 7.1.6.1 The cv-qualifiers paragraph 4 which says:

[...]any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.

Undefined behavior means the results are unpredictable which effectively means any result is possible even ones that at first glance defy intuition.

A quick experiment with godbolt using -O0 so there is no optimization going on shows the compiler is just using the literal value 15 for a instead of retrieving it from memory and printing that out:

movl    $15, %esi   #,

So the compiler is performing constant folding since it assumes that since a is constant it can just use the value 15 anywhere it sees a. Which is completely reasonable since you told it a was constant.

mechanism for "undefined-ness" of modifying the value of a const

1 Answers1