How does the validity of pointers interacts with uninitialized pointer value?

Question

I have seen that in the C++ abstract machines (if the are different), the mere act of forming a point to an invalid point in memory is undefined behavior.

For example

int* arr = new int[10];
int* last = arr + 9;  // last element
int* end = arr + 10; // past last element, still ok according to the rules
int* another = arr + 11;  // invalid, UB, supposedly, the rest of the program can be invalid

Now suppose, this other code

int* arr;  // arr can be invalid already here, UB?
int* another = arr;  // forming a pointer to possibly invalid memory, UB?

How is assigning an uninitialized pointer different from forming an invalid pointer by an operation? is it as bad?

int *arr; by itself is legal, int *another = arr; I think stops being UB in C++26? — HolyBlackCat
– HolyBlackCat, Commented Apr 27, 2024 at 5:24
C/C++ doesn't exist its either "C" or "C++" and in C++ using raw pointers is NOT recommended (specifically to avoid dereferencing nullptr's). In C++ you should use std::vector<int> in cases like this. std::vector<int> values; and then first and last will be values.begin() and values.end(). So be very very clear if you want a "C" or a "C++" answer ;) — Pepijn Kramer
– Pepijn Kramer, Commented Apr 27, 2024 at 5:28
Assigning a nullptr to another pointers isn't UB, dereferencing a nullptr is! So all of your code is fine (as long as you don't do anything with the pointers you created) — Pepijn Kramer
– Pepijn Kramer, Commented Apr 27, 2024 at 5:29
@PepijnKramer, yes, I meant two different abstract machines. Are they different in concrete ways? (as much as anything abstract can be concrete). I think for the question the same rules apply to both. fixed. — alfC
– alfC, Commented Apr 27, 2024 at 5:30

CPlus · Accepted Answer · 2024-04-27 05:27:16Z

1

int* arr;  // arr can be invalid already here, UB?

Not UB, since here you are not using/accessing the value of arr.

int* another = arr;  // forming a pointer to possibly invalid memory, UB?

UB, since you are accessing an indeterminate value to assign/initialize to another.

How is assigning an uninitialized pointer different from forming an invalid pointer by an operation? is it as bad?

They are both technically undefined behavior, so from a language lawyer perspective, they are equally bad, as they permit arbitrary behavior on behalf of the implementation.

answered Apr 27, 2024 at 5:27

CPlus

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4 Comments

alfC Over a year ago

Interesting! that means that any class containing a pointer (e.g., a pointer and a size) will have to have the pointer initialized (e.g., nullptr) for the class to have the assignment well-defined after default trivial construction.

CPlus Over a year ago

@alfC Not necessarily, as long as you do not access the pointer member directly. If I understand correctly, assignment of a structure (or class) to another is well defined, even if one of the members is indeterminate, i.e., accessing the member directly would have UB. Though admittedly, that Q&A is in C, not C++.

Pepijn Kramer Over a year ago

No member pointers do NOT have to be initialized (they aren't different in any way to normal pointers). Just as long as you have them initialized before first use (first dereference).

alfC Over a year ago

@CPlus, I think your linked question is a completely different beast since it doesn't involve a pointer

Richard · Accepted Answer · 2024-04-27 07:37:58Z

1

How is assigning an uninitialized pointer different from forming an invalid pointer by an operation?

In terms of initialization the definition int *arr; is no different than int i;. Both leave the variable uninitialized(assuming these appear in local/block scope and not in global scope).

The important thing is these definitions don't odr-use the variables arr and i.

On the other hand, when you do int *another = arr; you're using unitialized arr to initialize another. This constitutes an odr-use of uninitialized arr for something other than assignment or taking its address. And this is undefined behavior.

answered Apr 27, 2024 at 7:37

Richard

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4 Comments

Jeff Garrett Over a year ago

odr-use instead of lvalue-to-rvalue conversion?

Richard Over a year ago

@JeffGarrett Both I think.

alfC Over a year ago

@JeffGarrett, what is "lvalue-to-rvalue conversion" in this context?

Richard Over a year ago

@JeffGarrett As alfC mentioned/asked and I said, in int *arr; is no odr-use or lvalue-to-rvalue conversion.

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