Is the value of a variable always guaranteed to be 1 word away from the variable created outside of scope if it comes immediately after?

Question

int main()
{
    {int x = 532;}
    int xc;
    int x = *(&xc+1);
    cout << x << endl; // prints 532
}

I create an int variable with a value of 532 and then it immediately goes out of scope. I then create another int variable right after so that it has an address of 1 word before x. I then assign that value of whatever was one word past xc to x and print it, and it has the value of 532. Is this guaranteed to happen? Why/why not?

I get zero at optimization level 0, and 41988848 at optimization level 1, so clearly there is no guarantee. — Raymond Chen

paddy paddy · Accepted Answer · 2022-07-29T00:23:34

No, the arrangement and padding (and even the existence) of variables on the stack is implementation-dependent, and what you have here is Undefined Behavior. The language specifies that you must not use a pointer to access the memory of any object other than the one it points to.

In this case, &xc + 1 points to memory that lies outside the allocation of xc. You cannot read or write to this location and expect any kind of predictable result. Note that it is actually a valid pointer -- you are allowed to point "one-past-the-end" of an allocation, provided you never dereference that address.

To explain what's actually happening in your case: your variables are being pushed onto the stack, and the stack memory extends in the direction of lower addresses. That said, even this program on your computer with your compiler does not guarantee this behavior. The original variable x could be optimized away by the compiler completely because it is not used. The compiler is not required to understand that you're aliasing x via xc because, as already mentioned such behavior is not defined.

Is the value of a variable always guaranteed to be 1 word away from the variable created outside of scope if it comes immediately after?

1 Answers