Allocation function and pointer to returned

Question

I not understood suitable pointer alignment concept:

There are no constraints on the contents of the allocated storage on return from the allocation function. The order, contiguity, and initial value of storage allocated by successive calls to an allocation function are unspecified. The pointer returned shall be suitably aligned so that it can be converted to a pointer of any complete object type with a fundamental alignment requirement (3.11) and then used to access the object or array in the storage allocated (until the storage is explicitly deallocated by a call to a corresponding deallocation function).

There is no definition of suitable alignment in the sec.3.11. You explain that should mean?

If you're asking something about specific standard quotes, tagging with both C and C++ isn't really helpful. For C: _aligned_ is defined. — mafso, Jul 18 '14 at 18:23
Look here: http://stackoverflow.com/questions/8752546/how-does-malloc-understand-alignment — mafso, Jul 18 '14 at 18:24
You're jumping down a rabbit hole here. This question is not easily answered without producing a wall of text. Additionally, it is fairly unusual for most programmers to need to understand it. Some context might help us. Why do you want to know what alignment is? — John Dibling, Jul 18 '14 at 18:24
[What exactly is an aligned pointer?](http://stackoverflow.com/questions/4322926/what-exactly-is-an-aligned-pointer) — Deduplicator, Jul 18 '14 at 18:32

score 3 · Answer 1 · answered Jul 18 '14 at 18:30

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§3.11/1 says,

Object types have alignment requirements (3.9.1, 3.9.2) which place restrictions on the addresses at which an object of that type may be allocated.

So if a pointer is "suitably aligned" it means that the address represented by the pointer satisfies these restrictions. What exactly this means for the numerical value of the address is implementation-defined.

answered Jul 18 '14 at 18:30

Brian Bi

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That is, we can delete _suitable_ word from _suitable aligned_ without lost of meaning. – St.Antario Jul 18 '14 at 18:35
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@St.Antario I suppose so, yeah. – Brian Bi Jul 18 '14 at 18:36

score 3 · Accepted Answer · answered Jul 18 '14 at 18:35

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This means that for any complete object type with a fundamental alignment, it should be possible to convert the pointer returned to a pointer to that object type, respecting the alignment requirement of that object type.

In practice, since alignments are powers of two, this means that an allocation function is required to return a pointer aligned to alignof(std::max_align_t).

There is no separate definition of "suitable alignment"; in this paragraph as elsewhere "suitably" just means that there is a requirement which the program is required to satisfy for the rest of the paragraph to hold.

answered Jul 18 '14 at 18:35

ecatmur

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alignof(std::max_align_t)=16, but alignof(T*)=8 for any type T. – St.Antario Jul 18 '14 at 18:37
_There is no separate definition of "suitable alignment"_ That is all what I want to know! Thanks :). – St.Antario Jul 18 '14 at 18:39

score 2 · Answer 3 · answered Jul 18 '14 at 18:24

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Alignment is defined by the OS and platform. Usually it is the size of the largest basic type (a pointer or double), but it could be more. For instance, on Windows, x86 is 8 byte and x64 is 16 byte.

answered Jul 18 '14 at 18:24

Khouri Giordano

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It's worth noting that the largest fundamental type on x86/x86_64 is actually `long double`, which is 10 bytes in size. – Brian Bi Jul 18 '14 at 18:36
In MSVC 32 and 64-bit code, long double maps to double. – Khouri Giordano Jul 18 '14 at 18:39

Allocation function and pointer to returned

3 Answers3