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:

Answer 1

The C aliasing rules from which the rules of C++ were derived included a footnote specifying that the purpose of the rules was to say when things may alias. The authors of the Standard didn't think it necessary to forbid implementations from applying the rules in needlessly restrictive fashion in cases where things don't alias, because they thought compiler writers would honor the proverb "Don't prevent the programmer from doing what needs to be done", which the authors of the Standard viewed as part of the Spirit of C.

Situations where it would be necessary to use an lvalue of an aggregate's member type to actually alias a value of the aggregate type are rare, so it's entirely reasonable that the Standard doesn't require compilers to recognize such aliasing. Applying the rules restrictively in cases that don't involve aliasing, however, would cause something like:

union foo {int x; float y;} foo;
int *p = &foo.x;
*p = 1;

or even, for that matter,

union foo {int x; float y;} foo;
foo.x = 1;

to invoke UB since the assignment is used to access the stored values of a union foo and a float using an int, which is not one of the allowed types. Any quality compiler, however, should be able to recognize that an operation done on an lvalue which is visibly freshly derived from a union foo is an access to a union foo, and an access to a union foo is allowed to affect the stored values of its members (like the float member in this case).

The authors of the Standard probably declined to make the footnote normative because doing so would require a formal definition of when an access via freshly-derived lvalue is an access to the parent, and what kinds of access patterns constitute aliasing. While most cases would be pretty clear cut, there are some corner cases which implementations intended for low-level programming should probably interpret more pessimistically than those intended for e.g. high-end number crunching, and the authors of the Standard figured that anyone who could figure out how to handle the harder cases should be able to handle the easy ones.

Answer 2

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.