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Subject: Re: [ub] type punning through congruent base class?
From: Gabriel Dos Reis (gdr_at_[hidden])
Date: 2014-01-18 20:54:10


| -----Original Message-----
| From: ub-bounces_at_[hidden] [mailto:ub-bounces_at_[hidden]] On
| Behalf Of Kazutoshi Satoda
| Sent: Saturday, January 18, 2014 10:23 AM
| To: WG21 UB study group
| Cc: Bjarne Stroustrup
| Subject: Re: [ub] type punning through congruent base class?
|
| On 2014/01/17 13:58 +0900, Gabriel Dos Reis wrote:
| > Kazutoshi Satoda <k_satoda_at_[hidden]> writes:
| > | On 2014/01/17 5:53 +0900, Gabriel Dos Reis wrote:
| > | > On 2014/01/17 5:01 +0900, James Dennett wrote:
| > | > | struct B { int x };
| > | > | struct B* p = (B*) malloc(sizeof(B));
| > | > | p->x = 17;
| > | > |
| > | > | This (modulo the cast) has been how C has handled dynamic allocation
| > | > | of structs approximately forever. There are no constructors in C,
| > | > | structs don't get initialized, their fields just get assigned to.
| (snip)
| > | "p->x = 17" makes the effective type of the object between &p->x and
| > | (&p->x + 1) becomes int (6.5 p6), and stores value 17 (a value
| > | representation of int which represents 17) into that object (6.5.16).
| >
| > Agreed. In particular, at this point, we still do not have an object of
| > type B -- from C's perspective.
| >
| > | (Note)
| > | In C, there is no such a thing like "an object of type int" unlike in
| > | C++ where an object has a type and the type is determined when the
| > | object is created. In C, "object" is a merely a region of data storage,
| > | and may be labeled by an effective type which is determined by ongoing
| > | or previous access to the object at a time.
| >
| > Yes. This goes back to the comment I made:
| >
| > # Even if we take the C model, it is not clear what is there. We can
| > # infer that after the statement 'p->x = 17;' the storage at address
| > # &p-x' has effective type 'int', but that does not say much about the
| > # rest. I -suspect- C is OK with that because fundamentally, its object
| > # model is structural.
|
| There is an example which (I think) illustrates a real-world problem
| caused by this assertion "the effective type of the object is just int
| and not B." Please see this code and the result with gcc 4.8.2 (-O2).
| http://melpon.org/wandbox/permlink/BJQkgZDmbsHEp71j
|
| struct A { int a; };
| struct B { int a; double b; };
|
| int f(struct A* a, struct B* b)
| {
| b->a = 123;
| a->a = 456;
| return b->a;
| }
|
| Here, the compiler performs an optimization and returns constant 123
| from f(), assuming that two unrelated struct types never alias even if
| they have common initial members. I think this optimization is intended
| to be allowed by the aliasing rule in both C and C++.

Yes, I think this form of optimization -- type-based alias analysis -- is intended to be allowed in C++ and we definitely want to retain it. C's rules for type equivalence and type alias are on occasions a bit obscure for me to penetrate properly without tripping over. C11 fixed several bugs related to type aliasing and lifetime, but I don't know whether all C compilers were updated to the latest rules.

>From the C++ side, I see at least two issues with your program (but I think they are conceptually the same.)
The return statement performs a *read* using a path "b->a" that isn't valid under 3.10/10 (that you cite below). I would expect the use of this path in a write access to be OK, but I wouldn't expect it to OK for a read before the object construction is finished since there is no dynamic type set yet.

| But to allow this optimization in the above case, it is required to be
| OK that a compiler can assume the type of object at b is B, and the type
| of object at a is A, provided just the two assignment and one read to
| their member.

I'm unsure about that necessity. Remember that in C++ the dynamic type of an object isn't set before the object construction is complete. Yet we want this analysis to be applicable even during object construction.

| Then, it seems a de facto rule that an evaluation of member access
| operator solely (even when no actual access follow) implies the dynamic
| type (effective type in C) of the object designated by the left operand
| ("the object expression" in C++ term) to be a type which satisfies the
| aliasing rule as if the object is accessed by the static type of the
| object expression.

Not, necessarily. See my comment above. For C++, we know for sure the dynamic type isn't set before the constructor completes execution.

| Is this rule acceptable to be in the standard?

I am uncomfortable with that rule. We are trying to bring the essential common part of C and C++ regarding object lifetime to be in line with C++'s general object model.

| For C++, it would be like
| that:
|
| Change 3.10 p10 (C++) from:
| If a program attempts to access the stored value of an object
| through a glvalue of other than one of the following types the
| behavior is undefined:
| to
| If a program attempts to access the stored value of an object
| <ins>or evaluates a member access expression of non-static member
| (including interpretation of overloaded operators)</ins> through
| a glvalue of other than one of the following types the behavior is
| undefined:
|
| And define "reuse" to include (along with assignment to a glvalue of a
| scalar type, memcpy, etc) an evaluation of member access expression for
| trivially constructible class types, to take such a evaluation as a
| start of object lifetime with a certain type.
|
| With these rules, the above optimization is said to be OK because if
| "a->a = 456" was reusing the storage at b, the following access "return
| b->a" is undefined as it read from an object of type B which was killed
| by the reuse. Thus a compiler can assume that "a->a = 456" is not
| reusing b.

I think we both appear to agree that the read "b->a" is problematic. It looks like we are not quite agreeing on what makes it problematic. My view is that it is not OK because the object's dynamic type isn't defined so one can't use the path "b->a" to read the stored value (that is in line with 3.10/10).

The C equivalent would be that the effective type of the object at address 'b' isn't known so the read would trigger undefined behavior. But, as I said, C's type rules are trippy.

| (going back to analysis in C)
| > | If "struct B x = *p" follows the above code, the access (lvalue
| > | conversion, 6.3.2.1 p2) on *p, which is allowed by the aliasing rule
| > | (6.5 p7, bullet 5 "an aggregate or union ..."), changes the effective
| > | type of the object at p.
| (snip)
| > Bonus points: after
| >
| > struct B x = *p;
| >
| > what is the effective type of the object at p and &p->x? B and int or B
| > and B? Can we even give a meaning to the question?
| ...(from my follow up)
| > | Uh, sorry. I was wrong here. Non-modifying access to an object doesn't
| > | change the effective type of subsequent access to the object. And the
| > | aliasing rule does not apply here.
| >
| > Aha, but what about 6.5/6
| >
| > [...] For all other accesses to an object having no declared type,
| > the effective type of the object is simply the type of the lvalue use
| > for the access.
| >
| > ?
|
| A storing access determines the effective type for subsequent
| non-modifying accesses. And such subsequent non-modifying accesses do
| not drop into "all other accesses", I think.

Hmm, that is interesting and an interpretation I didn't consider earlier; but why would not they fall under "all other accesses"?

| Then, after "struct B x = *p", the effective type of the object at p
| and &p->x are both still int.

-- Gaby


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