I had a discussion with someone on IRC and this question turned up. We are allowed by the Standard to change an object of type int by a char lvalue.
int a;
char *b = (char*) &a;
*b = 0;
Would we be allowed to do this in the opposite direction, if we know that the alignment is fine?
The issue I'm seeing is that the aliasing rule does not cover the simple case of the following, if one considers the aliasing rule as a non-symmetric relation
int a;
a = 0;
The reason is, that each object contains a sequence of sizeof(obj) unsigned char objects (called the "object representation"). If we change the int, we will change some or all of those objects. However, the aliasing rule only states we are allowed to change a int by an char or unsigned char, but not the other way around. Another example
int a[1];
int *ra = a;
*ra = 0;
Only one direction is described by 3.10/15 ("An aggregate or union type that includes..."), but this time we need the other way around ("A type that is the element or non-static data member type of an aggregate...").
Is the other direction implied? This question also applies to C.
The aliasing rule simply states that there's one "effective type" (C99 6.5.7, plus footnote 73) for any given object in memory, and any accesses to such an object go through one of:
A type compatible with the effective type (qualifiers such as const and restrict, as well signed/unsigned-ness may vary)
A struct or union containing one such type
A character type
The effective type isn't specified in advanced, of course - it's just a construct that is used to specify aliasing. But the intent is simply that you don't access the same object with two different non-character types.
So the answer is, yes, you can indeed go the other direction.
The standard (C99 6.3.2.3 §7) defines pointer casts like this as "just fine", the casted pointer will point at the same address. (Unless the CPU has alignment which makes the cast impossible, then it is undefined behavior.)
That is, the actual cast in itself is fine. What will happen if you start to manipulate the data... now that's another implementation-defined story.
Here's from the standard:
"A pointer to an object or incomplete type may be converted to a pointer to a different object or incomplete type. If the resulting pointer is not correctly aligned (57) for the pointed-to type, the behavior is undefined. Otherwise, when converted back again, the result shall compare equal to the original pointer.
When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. Successive increments of the result, up to the size of the object, yield pointers
to the remaining bytes of the object."
"57) In general, the concept ‘‘correctly aligned’’ is transitive: if a pointer to type A is correctly aligned for a pointer to type B, which in turn is correctly aligned for a pointer to type C, then a pointer to type A is
correctly aligned for a pointer to type C."
I think I'm a bit confused by the question, but the 2nd and 3rd examples are accessing the int through an lvalue that has the object's type (int in the examples).
C++ 3.10/15 states as it's first item that it's OK to access the object through an lvalue that has the the type of "the dynamic type of the object".
What am I misunderstanding in the question?
Related
This question is a follow-up to: Is adding to a "char *" pointer UB, when it doesn't actually point to a char array?
In CWG 1314, CWG affirmed that it is legal to perform pointer arithmetic within a standard-layout object using an unsigned char pointer. This would appear to imply that some code similar to that in the linked question should work as intended:
struct Foo {
float x, y, z;
};
Foo f;
unsigned char *p = reinterpret_cast<unsigned char*>(&f) + offsetof(Foo, z); // (*)
*reinterpret_cast<float*>(p) = 42.0f;
(I have replaced char with unsigned char for greater clarity.)
However, it seems that the new changes in C++17 imply that this code is now UB unless std::launder is used after both reinterpret_casts. The result of a reinterpret_cast between two pointer types is equivalent to two static_casts: the first to cv void*, the second to the destination pointer type. But [expr.static.cast]/13 implies that this produces a pointer to the original object, not to an object of the destination type, since an object of type Foo is not pointer-interconvertible with an unsigned char object at its first byte, nor is an unsigned char object at the first byte of f.z pointer-interconvertible with f.z itself.
I find it hard to believe that the committee intended a change that would break this very common idiom, making all pre-C++17 usages of offsetof undefined.
You question was:
Do we need to use std::launder when doing pointer arithmetic within a
standard-layout object (e.g., with offsetof)?
No.
std::launder won't change anything in this case and therefore has nothing to do with the presented example (imo edit launder out of the question or ask another question).
std::launder is usually just needed in a subset of cases (eg. due to a const member) where you change (or create) an underlying object in some runtime manner (eg. via placement new). Mnemonic: the object is 'dirty' and needs to be std::launder'ed.
Using only a standard layout type cannot result in a situation where you would ever need to use std::launder.
The current draft standard (and presumably C++17) say in [basic.compound/4]:
[ Note: An array object and its first element are not pointer-interconvertible, even though they have the same address. — end note ]
So a pointer to an object cannot be reinterpret_cast'd to get its enclosing array pointer.
Now, there is std::launder, [ptr.launder/1]:
template<class T> [[nodiscard]] constexpr T* launder(T* p) noexcept;
Requires: p represents the address A of a byte in memory. An object X that is within its lifetime and whose type is similar to T is located at the address A. All bytes of storage that would be reachable through the result are reachable through p (see below).
And the definion of reachable is in [ptr.launder/3]:
Remarks: An invocation of this function may be used in a core constant expression whenever the value of its argument may be used in a core constant expression. A byte of storage is reachable through a pointer value that points to an object Y if it is within the storage occupied by Y, an object that is pointer-interconvertible with Y, or the immediately-enclosing array object if Y is an array element. The program is ill-formed if T is a function type or cv void.
Now, at first sight, it seems that std::launder is can be used to do the aforementioned conversion, because of the part I've put emphasis.
But. If p points to an object of an array, the bytes of the array is reachable according to this definition (even though p is not pointer-interconvertible to array-pointer), just like the result of the launder. So, it seems that the definition doesn't say anything about this issue.
So, can std::launder be used to convert an object pointer to its enclosing array pointer?
This depends on whether the enclosing array object is a complete object, and if not, whether you can validly access more bytes through a pointer to that enclosing array object (e.g., because it's an array element itself, or pointer-interconvertible with a larger object, or pointer-interconvertible with an object that's an array element). The "reachable" requirement means that you cannot use launder to obtain a pointer that would allow you to access more bytes than the source pointer value allows, on pain of undefined behavior. This ensures that the possibility that some unknown code may call launder does not affect the compiler's escape analysis.
I suppose some examples could help. Each example below reinterpret_casts a int* pointing to the first element of an array of 10 ints into a int(*)[10]. Since they are not pointer-interconvertible, the reinterpret_cast does not change the pointer value, and you get a int(*)[10] with the value of "pointer to the first element of (whatever the array is)". Each example then attempts to obtain a pointer to the entire array by calling std::launder on the cast pointer.
int x[10];
auto p = std::launder(reinterpret_cast<int(*)[10]>(&x[0]));
This is OK; you can access all elements of x through the source pointer, and the result of the launder doesn't allow you to access anything else.
int x2[2][10];
auto p2 = std::launder(reinterpret_cast<int(*)[10]>(&x2[0][0]));
This is undefined. You can only access elements of x2[0] through the source pointer, but the result (which would be a pointer to x2[0]) would have allowed you to access x2[1], which you can't through the source.
struct X { int a[10]; } x3, x4[2]; // assume no padding
auto p3 = std::launder(reinterpret_cast<int(*)[10]>(&x3.a[0])); // OK
This is OK. Again, you can't access through a pointer to x3.a any byte you can't access already.
auto p4 = std::launder(reinterpret_cast<int(*)[10]>(&x4[0].a[0]));
This is (intended to be) undefined. You would have been able to reach x4[1] from the result because x4[0].a is pointer-interconvertible with x4[0], so a pointer to the former can be reinterpret_cast to yield a pointer to the latter, which then can be used for pointer arithmetic. See https://wg21.link/LWG2859.
struct Y { int a[10]; double y; } x5;
auto p3 = std::launder(reinterpret_cast<int(*)[10]>(&x5.a[0]));
And this is again undefined, because you would have been able to reach x5.y from the resulting pointer (by reinterpret_cast to a Y*) but the source pointer can't be used to access it.
Remark: any non schizophrenic compiler will probably gladly accept that, as it would accept a C-style cast or a re-interpret cast, so just try and see is not an option.
But IMHO, the answer to your question is no. The emphasized immediately-enclosing array object if Y is an array element lies in a Remark paragraph, not in the Requires one. That means that provided the requires section is respected, the remarks one also applies. As an array and its element type are not similar types, the requirement is not satisfied and std::launder cannot be used.
What follows is more of a general (philosophycal?) interpretation. At the time of K&R C (in the 70's), C was intended to be able to replace assembly language. For that reason the rule was: the compiler must obey the programmer provided the source code can be translated. So no strict aliasing rule and a pointer was no more that an address with additional arithmetics rules. This strongly changed in C99 and C++03 (not speaking of C++11 +). Programmers are now supposed to use C++ as a high level language. That means that a pointer is just an object that allows to access another object of a given type, and an array and its element type are totally different types. Memory addresses are now little more than implementation details. So trying to convert a pointer to an array to a pointer to its first element is then against the philosophy of the language and could bite the programmer in a later version of the compiler. Of course real life compiler still accept it for compatibility reasons, but we should not even try to use it in modern programs.
The current draft standard (and presumably C++17) say in [basic.compound/4]:
[ Note: An array object and its first element are not pointer-interconvertible, even though they have the same address. — end note ]
So a pointer to an object cannot be reinterpret_cast'd to get its enclosing array pointer.
Now, there is std::launder, [ptr.launder/1]:
template<class T> [[nodiscard]] constexpr T* launder(T* p) noexcept;
Requires: p represents the address A of a byte in memory. An object X that is within its lifetime and whose type is similar to T is located at the address A. All bytes of storage that would be reachable through the result are reachable through p (see below).
And the definion of reachable is in [ptr.launder/3]:
Remarks: An invocation of this function may be used in a core constant expression whenever the value of its argument may be used in a core constant expression. A byte of storage is reachable through a pointer value that points to an object Y if it is within the storage occupied by Y, an object that is pointer-interconvertible with Y, or the immediately-enclosing array object if Y is an array element. The program is ill-formed if T is a function type or cv void.
Now, at first sight, it seems that std::launder is can be used to do the aforementioned conversion, because of the part I've put emphasis.
But. If p points to an object of an array, the bytes of the array is reachable according to this definition (even though p is not pointer-interconvertible to array-pointer), just like the result of the launder. So, it seems that the definition doesn't say anything about this issue.
So, can std::launder be used to convert an object pointer to its enclosing array pointer?
This depends on whether the enclosing array object is a complete object, and if not, whether you can validly access more bytes through a pointer to that enclosing array object (e.g., because it's an array element itself, or pointer-interconvertible with a larger object, or pointer-interconvertible with an object that's an array element). The "reachable" requirement means that you cannot use launder to obtain a pointer that would allow you to access more bytes than the source pointer value allows, on pain of undefined behavior. This ensures that the possibility that some unknown code may call launder does not affect the compiler's escape analysis.
I suppose some examples could help. Each example below reinterpret_casts a int* pointing to the first element of an array of 10 ints into a int(*)[10]. Since they are not pointer-interconvertible, the reinterpret_cast does not change the pointer value, and you get a int(*)[10] with the value of "pointer to the first element of (whatever the array is)". Each example then attempts to obtain a pointer to the entire array by calling std::launder on the cast pointer.
int x[10];
auto p = std::launder(reinterpret_cast<int(*)[10]>(&x[0]));
This is OK; you can access all elements of x through the source pointer, and the result of the launder doesn't allow you to access anything else.
int x2[2][10];
auto p2 = std::launder(reinterpret_cast<int(*)[10]>(&x2[0][0]));
This is undefined. You can only access elements of x2[0] through the source pointer, but the result (which would be a pointer to x2[0]) would have allowed you to access x2[1], which you can't through the source.
struct X { int a[10]; } x3, x4[2]; // assume no padding
auto p3 = std::launder(reinterpret_cast<int(*)[10]>(&x3.a[0])); // OK
This is OK. Again, you can't access through a pointer to x3.a any byte you can't access already.
auto p4 = std::launder(reinterpret_cast<int(*)[10]>(&x4[0].a[0]));
This is (intended to be) undefined. You would have been able to reach x4[1] from the result because x4[0].a is pointer-interconvertible with x4[0], so a pointer to the former can be reinterpret_cast to yield a pointer to the latter, which then can be used for pointer arithmetic. See https://wg21.link/LWG2859.
struct Y { int a[10]; double y; } x5;
auto p3 = std::launder(reinterpret_cast<int(*)[10]>(&x5.a[0]));
And this is again undefined, because you would have been able to reach x5.y from the resulting pointer (by reinterpret_cast to a Y*) but the source pointer can't be used to access it.
Remark: any non schizophrenic compiler will probably gladly accept that, as it would accept a C-style cast or a re-interpret cast, so just try and see is not an option.
But IMHO, the answer to your question is no. The emphasized immediately-enclosing array object if Y is an array element lies in a Remark paragraph, not in the Requires one. That means that provided the requires section is respected, the remarks one also applies. As an array and its element type are not similar types, the requirement is not satisfied and std::launder cannot be used.
What follows is more of a general (philosophycal?) interpretation. At the time of K&R C (in the 70's), C was intended to be able to replace assembly language. For that reason the rule was: the compiler must obey the programmer provided the source code can be translated. So no strict aliasing rule and a pointer was no more that an address with additional arithmetics rules. This strongly changed in C99 and C++03 (not speaking of C++11 +). Programmers are now supposed to use C++ as a high level language. That means that a pointer is just an object that allows to access another object of a given type, and an array and its element type are totally different types. Memory addresses are now little more than implementation details. So trying to convert a pointer to an array to a pointer to its first element is then against the philosophy of the language and could bite the programmer in a later version of the compiler. Of course real life compiler still accept it for compatibility reasons, but we should not even try to use it in modern programs.
Note: This question has been renamed and reduced to make it more focused and readable. Most of the comments refer to the old text.
According to the standard, objects of different type may not share the same memory location. So this would not be legal:
std::array<short, 4> shorts;
int* i = reinterpret_cast<int*>(shorts.data()); // Not OK
The standard, however, allows an exception to this rule: any object may be accessed through a pointer to char or unsigned char:
int i = 0;
char * c = reinterpret_cast<char*>(&i); // OK
However, it is not clear to me whether this is also allowed the other way around. For example:
char * c = read_socket(...);
unsigned * u = reinterpret_cast<unsigned*>(c); // huh?
Some of your code is questionable due to the pointer conversions involved. Keep in mind that in those instances reinterpret_cast<T*>(e) has the semantics of static_cast<T*>(static_cast<void*>(e)) because the types that are involved are standard-layout. (I would in fact recommend that you always use static_cast via cv void* when dealing with storage.)
A close reading of the Standard suggests that during a pointer conversion to or from T* it is assumed that there really is an actual object T* involved -- which is hard to fulfill in some of your snippet, even when 'cheating' thanks to the triviality of types involved (more on this later). That would be besides the point however because...
Aliasing is not about pointer conversions. This is the C++11 text that outlines the rules that are commonly referred to as 'strict aliasing' rules, from 3.10 Lvalues and rvalues [basic.lval]:
10 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:
the dynamic type of the object,
a cv-qualified version of the dynamic type of the object,
a type similar (as defined in 4.4) to the dynamic type of the object,
a type that is the signed or unsigned type corresponding to the dynamic type of the object,
a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic type of the object,
an aggregate or union type that includes one of the aforementioned types among its elements or non-static data members (including, recursively, an element or non-static data member of a subaggregate or contained union),
a type that is a (possibly cv-qualified) base class type of the dynamic type of the object,
a char or unsigned char type.
(This is paragraph 15 of the same clause and subclause in C++03, with some minor changes in the text with e.g. 'lvalue' being used instead of 'glvalue' since the latter is a C++11 notion.)
In the light of those rules, let's assume that an implementation provides us with magic_cast<T*>(p) which 'somehow' converts a pointer to another pointer type. Normally this would be reinterpret_cast, which yields unspecified results in some cases, but as I've explained before this is not so for pointers to standard-layout types. Then it's plainly true that all of your snippets are correct (substituting reinterpret_cast with magic_cast), because no glvalues are involved whatsoever with the results of magic_cast.
Here is a snippet that appears to incorrectly use magic_cast, but which I will argue is correct:
// assume constexpr max
constexpr auto alignment = max(alignof(int), alignof(short));
alignas(alignment) char c[sizeof(int)];
// I'm assuming here that the OP really meant to use &c and not c
// this is, however, inconsequential
auto p = magic_cast<int*>(&c);
*p = 42;
*magic_cast<short*>(p) = 42;
To justify my reasoning, assume this superficially different snippet:
// alignment same as before
alignas(alignment) char c[sizeof(int)];
auto p = magic_cast<int*>(&c);
// end lifetime of c
c.~decltype(c)();
// reuse storage to construct new int object
new (&c) int;
*p = 42;
auto q = magic_cast<short*>(p);
// end lifetime of int object
p->~decltype(0)();
// reuse storage again
new (p) short;
*q = 42;
This snippet is carefully constructed. In particular, in new (&c) int; I'm allowed to use &c even though c was destroyed due to the rules laid out in paragraph 5 of 3.8 Object lifetime [basic.life]. Paragraph 6 of same gives very similar rules to references to storage, and paragraph 7 explains what happens to variables, pointers and references that used to refer to an object once its storage is reused -- I will refer collectively to those as 3.8/5-7.
In this instance &c is (implicitly) converted to void*, which is one of the correct use of a pointer to storage that has not been yet reused. Similarly p is obtained from &c before the new int is constructed. Its definition could perhaps be moved to after the destruction of c, depending on how deep the implementation magic is, but certainly not after the int construction: paragraph 7 would apply and this is not one of the allowed situations. The construction of the short object also relies on p becoming a pointer to storage.
Now, because int and short are trivial types, I don't have to use the explicit calls to destructors. I don't need the explicit calls to the constructors, either (that is to say, the calls to the usual, Standard placement new declared in <new>). From 3.8 Object lifetime [basic.life]:
1 [...] The lifetime of an object of type T begins when:
storage with the proper alignment and size for type T is obtained, and
if the object has non-trivial initialization, its initialization is complete.
The lifetime of an object of type T ends when:
if T is a class type with a non-trivial destructor (12.4), the destructor call starts, or
the storage which the object occupies is reused or released.
This means that I can rewrite the code such that, after folding the intermediate variable q, I end up with the original snippet.
Do note that p cannot be folded away. That is to say, the following is defintively incorrect:
alignas(alignment) char c[sizeof(int)];
*magic_cast<int*>(&c) = 42;
*magic_cast<short*>(&c) = 42;
If we assume that an int object is (trivially) constructed with the second line, then that must mean &c becomes a pointer to storage that has been reused. Thus the third line is incorrect -- although due to 3.8/5-7 and not due to aliasing rules strictly speaking.
If we don't assume that, then the second line is a violation of aliasing rules: we're reading what is actually a char c[sizeof(int)] object through a glvalue of type int, which is not one of the allowed exception. By comparison, *magic_cast<unsigned char>(&c) = 42; would be fine (we would assume a short object is trivially constructed on the third line).
Just like Alf, I would also recommend that you explicitly make use of the Standard placement new when using storage. Skipping destruction for trivial types is fine, but when encountering *some_magic_pointer = foo; you're very much likely facing either a violation of 3.8/5-7 (no matter how magically that pointer was obtained) or of the aliasing rules. This means storing the result of the new expression, too, since you most likely can't reuse the magic pointer once your object is constructed -- due to 3.8/5-7 again.
Reading the bytes of an object (this means using char or unsigned char) is fine however, and you don't even to use reinterpret_cast or anything magic at all. static_cast via cv void* is arguably fine for the job (although I do feel like the Standard could use some better wording there).
This too:
// valid: char -> type
alignas(int) char c[sizeof(int)];
int * i = reinterpret_cast<int*>(c);
That is not correct. The aliasing rules state under which circumstances it is legal/illegal to access an object through an lvalue of a different type. There is an specific rule that says that you can access any object through a pointer of type char or unsigned char, so the first case is correct. That is, A => B does not necessarily mean B => A. You can access an int through a pointer to char, but you cannot access a char through a pointer to int.
For the benefit of Alf:
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:
the dynamic type of the object,
a cv-qualified version of the dynamic type of the object,
a type similar (as defined in 4.4) to the dynamic type of the object,
a type that is the signed or unsigned type corresponding to the dynamic type of the object,
a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic type of the object,
an aggregate or union type that includes one of the aforementioned types among its elements or non- static data members (including, recursively, an element or non-static data member of a subaggregate or contained union),
a type that is a (possibly cv-qualified) base class type of the dynamic type of the object,
a char or unsigned char type.
Regarding the validity of …
alignas(int) char c[sizeof(int)];
int * i = reinterpret_cast<int*>(c);
The reinterpret_cast itself is OK or not, in the sense of producing a useful pointer value, depending on the compiler. And in this example the result isn't used, in particular, the character array isn't accessed. So there is not much more that can be said about the example as-is: it just depends.
But let's consider an extended version that does touch on the aliasing rules:
void foo( char* );
alignas(int) char c[sizeof( int )];
foo( c );
int* p = reinterpret_cast<int*>( c );
cout << *p << endl;
And let's only consider the case where the compiler guarantees a useful pointer value, one that would place the pointee in the same bytes of memory (the reason that this depends on the compiler is that the standard, in §5.2.10/7, only guarantees it for pointer conversions where the types are alignment-compatible, and otherwise leave it as "unspecified" (but then, the whole of §5.2.10 is somewhat inconsistent with §9.2/18).
Now, one interpretation of the standard's §3.10/10, the so called "strict aliasing" clause (but note that the standard does not ever use the term "strict aliasing"),
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:
the dynamic type of the object,
a cv-qualified version of the dynamic type of the object,
a type similar (as defined in 4.4) to the dynamic type of the object,
a type that is the signed or unsigned type corresponding to the dynamic type of the object,
a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic type of the object,
an aggregate or union type that includes one of the aforementioned types among its elements or non- static data members (including, recursively, an element or non-static data member of a subaggregate or contained union),
a type that is a (possibly cv-qualified) base class type of the dynamic type of the object,
a char or unsigned char type.
is that, as it itself says, concerns the dynamic type of the object residing in the c bytes.
With that interpretation, the read operation on *p is OK if foo has placed an int object there, and otherwise not. So in this case, a char array is accessed via an int* pointer. And nobody is in any doubt that the other way is valid: even though foo may have placed an int object in those bytes, you can freely access that object as a sequence of char values, by the last dash of §3.10/10.
So with this (usual) interpretation, after foo has placed an int there, we can access it as char objects, so at least one char object exists within the memory region named c; and we can access it as int, so at least that one int exists there also; and so David’s assertion in another answer that char objects cannot be accessed as int, is incompatible with this usual interpretation.
David's assertion is also incompatible with the most common use of placement new.
Regarding what other possible interpretations there are, that perhaps could be compatible with David's assertion, well, I can't think of any that make sense.
So in conclusion, as far as the Holy Standard is concerned, merely casting oneself a T* pointer to the array is practically useful or not depending on the compiler, and accessing the pointed to could-be-value is valid or not depending on what's present. In particular, think of a trap representation of int: you would not want that blowing up on you, if the bitpattern happened to be that. So to be safe you have to know what's in there, the bits, and as the call to foo above illustrates the compiler can in general not know that, like, the g++ compiler's strict alignment-based optimizer can in general not know that…
According to strict aliasing rules:
struct B { virtual ~B() {} };
struct D : public B { };
D d;
char *c = reinterpret_cast<char*>(&d);
A char* to any object of different type is valid. But now the question is, will it point to the same address of &d? what is the guarantee made by C++ Standard that it will return the same address?
c and &d do indeed have the same value, and if you reinterpret-cast c back to a D* you get a valid pointer that you may dereference. Furthermore, you can treat c as (pointer to the first element of) an opaque array char[sizeof(D)] -- this is indeed the main purpose of casting pointers to char pointers: To allow (de)serialization (e.g. ofile.write(c, sizeof(D));), although you should generally only do this for primitive types (and arrays thereof), since the binary layout of of compound types is not generally specified in a portable fashion.
As #Oli rightly points out and would like me to reinforce, you should really never serialize compound types as a whole. The result will almost never be deserializable, since the implementation of polymorphic classes and padding between data fields is not specified and not accessible to you.
Note that reinterpret_cast<char*>(static_cast<B*>(&d)) may be treated as an opaque array char[sizeof(B)] by similar reasoning.
Section 5.2.10, point 7 of the 2003 C++ Standard says:
A pointer to an object can be explicitly converted to a pointer to an
object of different type. Except that converting an rvalue of type
“pointer to T1” to the type “pointer to T2” (where T1 and T2 are
object types and where the alignment requirements of T2 are no
stricter than those of T1) and back to its original type yields the
original pointer value, the result of such a pointer conversion is
unspecified.
If by "same address" you mean "original pointer value," then this entry says "yes."
The intent is clear (and not something that needs to be debated):
reinterpret_cast never changes the value of an address, unless the target type cannot represent all address values (like a small integer type, on a pointer type with intrinsic alignment: f.ex. a pointer that can only represent even addresses, or pointers to object and pointers to functions cannot be mixed...).
The wording of the standard fails to capture that, but that doesn't mean there is a real practical issue here.
char *c = reinterpret_cast<char*>(&d);
c will point to the first byte of d, always.