Strict aliasing rule

Question

I'm reading notes about reinterpret_cast and it's aliasing rules ( http://en.cppreference.com/w/cpp/language/reinterpret_cast ).

I wrote that code:

struct A
{
  int t;
};

char *buf = new char[sizeof(A)];

A *ptr = reinterpret_cast<A*>(buf);
ptr->t = 1;

A *ptr2 = reinterpret_cast<A*>(buf);
cout << ptr2->t;

I think these rules doesn't apply here:

T2 is the (possibly cv-qualified) dynamic type of the object
T2 and T1 are both (possibly multi-level, possibly cv-qualified at each level) pointers to the same type T3 (since C++11)
T2 is an aggregate type or a union type which holds one of the aforementioned types as an element or non-static member (including, recursively, elements of subaggregates and non-static data members of the contained unions): this makes it safe to cast from the first member of a struct and from an element of a union to the struct/union that contains it.
T2 is the (possibly cv-qualified) signed or unsigned variant of the dynamic type of the object
T2 is a (possibly cv-qualified) base class of the dynamic type of the object
T2 is char or unsigned char

In my opinion this code is incorrect. Am I right? Is code correct or not?

On the other hand what about connect function (man 2 connect) and struct sockaddr?

   int connect(int sockfd, const struct sockaddr *addr,
               socklen_t addrlen);

Eg. we have struct sockaddr_in and we have to cast it to struct sockaddr. Above rules also doesn't apply, so is this cast incorrect?

Make that char buf[sizeof(A)] and gcc will detect both violations at -Wstrict-aliasing=2 — Cubbi
Type aliasing rules are about access to stored objects, not about casts. Casting some pointer type to another pointer type can't break strict aliasing rules; you need to dereference a pointer to break the rules. — davmac
@Adam, yes, your code breaks the rules. But you ask about the connect method, and specifically - "Above rules also doesn't apply, so is this cast incorrect?" - as I say, the aliasing rules are not about casts. — davmac
Please note I have edited my answer to take into account alignment considerations (with which I wasn't overly familiar in 2015) — Lightness Races in Orbit

Lightness Races in Orbit Lightness Races in Orbit · Accepted Answer · 2015-07-24T16:19:34

Yeah, it's invalid, but not because you're converting a char* to an A*: it's because you are not obtaining a A* that actually points to an A* and, as you've identified, none of the type aliasing options fit.

You'd need something like this:

#include <new>
#include <iostream>

struct A
{
  int t;
};

char *buf = new char[sizeof(A)];

A* ptr = new (buf) A;
ptr->t = 1;

// Also valid, because points to an actual constructed A!
A *ptr2 = reinterpret_cast<A*>(buf);
std::cout << ptr2->t;

Now type aliasing doesn't come into it at all (though keep reading because there's more to do!).

(live demo with -Wstrict-aliasing=2)

In reality, this is not enough. We must also consider alignment. Though the above code may appear to work, to be fully safe and whatnot you will need to placement-new into a properly-aligned region of storage, rather than just a casual block of chars.

The standard library (since C++11) gives us std::aligned_storage to do this:

using Storage = std::aligned_storage<sizeof(A), alignof(A)>::type;
auto* buf = new Storage;

Or, if you don't need to dynamically allocate it, just:

Storage data;

Then, do your placement-new:

new (buf) A();
// or: new(&data) A();

And to use it:

auto ptr = reinterpret_cast<A*>(buf);
// or: auto ptr = reinterpret_cast<A*>(&data);

All in it looks like this:

#include <iostream>
#include <new>
#include <type_traits>

struct A
{
  int t;
};

int main()
{
    using Storage = std::aligned_storage<sizeof(A), alignof(A)>::type;

    auto* buf = new Storage;
    A* ptr = new(buf) A();

    ptr->t = 1;

    // Also valid, because points to an actual constructed A!
    A* ptr2 = reinterpret_cast<A*>(buf);
    std::cout << ptr2->t;
}

(live demo)

Even then, since C++17 this is somewhat more complicated; see the relevant cppreference pages for more information and pay attention to std::launder.

Of course, this whole thing appears contrived because you only want one A and therefore don't need array form; in fact, you'd just create a bog-standard A in the first place. But, assuming buf is actually larger in reality and you're creating an allocator or something similar, this makes some sense.

Strict aliasing rule

2 Answers

(live demo)