treating memory returned by operator new(sizeof(T) * N) as an array
The C++ standards contain an open issue that underlying representation of objects is not an "array" but a "sequence" of
unsigned char objects. Still, everyone treats it as an array (which is intended), so it is safe to write the code like:
char* storage = static_cast<char*>(operator new(sizeof(T)*size));
char* p = storage + sizeof(T)*i; // precondition: 0 <= i < size
new (p) T(element);
as long as
void* operator new(size_t) returns a properly aligned value. Using
sizeof-multiplied offsets to keep the alignment is safe.
In C++17, there is a macro STDCPP_DEFAULT_NEW_ALIGNMENT, which specifies the maximum safe alignment for "normal"
void* operator new(size_t), and
void* operator new(std::size_t size, std::align_val_t alignment) should be used if a larger alignment is required.
In earlier versions of C++, there is no such distinction, which means that
void* operator new(size_t) needs to be implemented in a way that is compatible with the alignment of any object.
As to being able to do pointer arithmetic directly on
T*, I am not sure it needs to be required by the standard. However, it is hard to implement the C++ memory model in such a way that it would not work.
Array placement-new requires unspecified overhead in the buffer?
Nicol Bolas correctly points out in the comments below that this has been fixed such that the overhead is always zero for
operator new(std::size_t, void* p).
This fix was done as a defect report in November 2019, which makes it retroactive to all versions of C++.
operator new(std::size_t, void* p) unless you know a-priori the answer to this question. The answer is an implementation detail and can change with compiler/platform. Though it is typically stable for any given platform. E.g. this is something specified by the Itanium ABI.
If you don't know the answer to this question, write your own placement array new that can check this at run time:
operator new(std::size_t n, void* p, std::size_t limit)
if (n <= limit)
std::cout << "life is good\n";
alignas(std::string) char buffer;
std::string* p = new(buffer, sizeof(buffer)) std::string;
By varying the array size and inspecting
n in the example above, you can infer
y for your platform. For my platform
y is 1 word. The sizeof(word) varies depending on whether I'm compiling for a 32 bit or 64 bit architecture.
How to create objects in C++ with dynamic alignment requirements?
You can do the following:
auto storage = static_cast<Foo*>(std::aligned_alloc(n * sizeof(Foo), alignment));
auto ptr = std::launder(storage);
// use ptr to refer to Foo objects
Casting a pointer returned by
Foo* is not undefined behavior. It would be UB if you tried to dereference it right after
The code above is technically undefined behavior. But it seems that in C++ there is no 100% standard-conforming way to do such an allocation without UB. From the practical point to view this code is reliable and safe.
std::launder(storage) should probably be used to access
Foo objects through
See this question for details and discussions.
Dynamic arrays in C++ without Undefined Behavior
std::vector can't, then you can't either.
But I wouldn't worry about it. This is one of those cases where people have found a problem with the wording of the standard, that technically makes an extremely common use case undefined. But your vectors still work.
Now, the key: that's not because of magic innate to
std::vector, or to some particular
std::vector implementation: it's because the compiler doesn't perform absurd optimisations that make use of this undefined behaviour that somebody only just spotted while studying the text with a fine-toothed comb.
Perhaps it'll be tidied up in a future revision, but for practical purposes you do not need to worry about it, whether you use
std::vector or you use
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