Determining the alignment of C/C++ structures in relation to its members - c++

Question

Can the alignment of a structure type be found if the alignments of the structure members are known?
Eg. for:
struct S
{
a_t a;
b_t b;
c_t c[];
};
is the alignment of S = max(alignment_of(a), alignment_of(b), alignment_of(c))?
Searching the internet I found that "for structured types the largest alignment requirement of any of its elements determines the alignment of the structure" (in What Every Programmer Should Know About Memory) but I couldn't find anything remotely similar in the standard (latest draft more exactly).
Edited:
Many thanks for all the answers, especially to Robert Gamble who provided a really good answer to the original question and the others who contributed.
In short:
To ensure alignment requirements for structure members, the alignment of a structure must be at least as strict as the alignment of its strictest member.
As for determining the alignment of structure a few options were presented and with a bit of research this is what I found:
c++ std::tr1::alignment_of
not standard yet, but close (technical report 1), should be in the C++0x
the following restrictions are present in the latest draft: Precondition:T shall be a complete type, a reference type, or an array of
unknown bound, but shall not be a function type or (possibly
cv-qualified) void.
this means that my presented use case with the C99 flexible array won't work (this is not that surprising since flexible arrays are not standard c++)
in the latest c++ draft it is defined in the terms of a new keyword - alignas (this has the same complete type requirement)
in my opinion, should c++ standard ever support C99 flexible arrays, the requirement could be relaxed (the alignment of the structure with the flexible array should not change based on the number of the array elements)
c++ boost::alignment_of
mostly a tr1 replacement
seems to be specialized for void and returns 0 in that case (this is forbidden in the c++ draft)
Note from developers: strictly speaking you should only rely on the value of ALIGNOF(T) being a multiple of the true alignment of T, although in practice it does compute the correct value in all the cases we know about.
I don't know if this works with flexible arrays, it should (might not work in general, this resolves to compiler intrinsic on my platform so I don't know how it will behave in the general case)
Andrew Top presented a simple template solution for calculating the alignment in the answers
this seems to be very close to what boost is doing (boost will additionally return the object size as the alignment if it is smaller than the calculated alignment as far as I can see) so probably the same notice applies
this works with flexible arrays
use Windbg.exe to find out the alignment of a symbol
not compile time, compiler specific, didn't test it
using offsetof on the anonymous structure containing the type
see the answers, not reliable, not portable with c++ non-POD
compiler intrinsics, eg. MSVC __alignof
works with flexible arrays
alignof keyword is in the latest c++ draft
If we want to use the "standard" solution we're limited to std::tr1::alignment_of, but that won't work if you mix your c++ code with c99's flexible arrays.
As I see it there is only 1 solution - use the old struct hack:
struct S
{
a_t a;
b_t b;
c_t c[1]; // "has" more than 1 member, strictly speaking this is undefined behavior in both c and c++ when used this way
};
The diverging c and c++ standards and their growing differences are unfortunate in this case (and every other case).
Another interesting question is (if we can't find out the alignment of a structure in a portable way) what is the most strictest alignment requirement possible. There are a couple of solutions I could find:
boost (internally) uses a union of variety of types and uses the boost::alignment_of on it
the latest c++ draft contains std::aligned_storage
The value of default-alignment shall be the most stringent alignment requirement for any C++ object type whose size is no greater than Len
so the std::alignment_of< std::aligned_storage<BigEnoughNumber>>::value should give us the maximum alignment
draft only, not standard yet (if ever), tr1::aligned_storage does not have this property
Any thoughts on this would also be appreciated.
I have temporarily unchecked the accepted answer to get more visibility and input on the new sub-questions

Answer 1

There are two closely related concepts to here:
The alignment required by the processor to access a particular object
The alignment that the compiler actually uses to place objects in memory
To ensure alignment requirements for structure members, the alignment of a structure must be at least as strict as the alignment of its strictest member. I don't think this is spelled out explicitly in the standard but it can be inferred from the the following facts (which are spelled out individually in the standard):
Structures are allowed to have padding between their members (and at the end)
Arrays are not allowed to have padding between their elements
You can create an array of any structure type
If the alignment of a structure was not at least as strict as each of its members you would not be able to create an array of structures since some structure members some elements would not be properly aligned.
Now the compiler must ensure a minimum alignment for the structure based on the alignment requirements of its members but it can also align objects in a stricter fashion than required, this is often done for performance reasons. For example, many modern processors will allow access to 32-bit integers in any alignment but accesses may be significantly slower if they are not aligned on a 4-byte boundary.
There is no portable way to determine the alignment enforced by the processor for any given type because this is not exposed by the language, although since the compiler obviously knows the alignment requirements of the target processor it could expose this information as an extension.
There is also no portable way (at least in C) to determine how a compiler will actually align an object although many compilers have options to provide some level of control over the alignment.

Answer 2

I wrote this type trait code to determine the alignment of any type(based on the compiler rules already discussed). You may find it useful:
template <class T>
class Traits
{
public:
struct AlignmentFinder
{
char a;
T b;
};
enum {AlignmentOf = sizeof(AlignmentFinder) - sizeof(T)};
};
So now you can go:
std::cout << "The alignment of structure S is: " << Traits<S>::AlignmentOf << std::endl;

Answer 3

The following macro will return the alignment requirement of any given type (even if it's a struct):
#define TYPE_ALIGNMENT( t ) offsetof( struct { char x; t test; }, test )
Note: I probably borrowed this idea from a Microsoft header at some point way back in my past...
Edit: as Robert Gamble points out in the comments, this macro is not guaranteed to work. In fact, it will certainly not work very well if the compiler is set to pack elements in structures. So if you decide to use it, use it with caution.
Some compilers have an extension that allows you obtain the alignment of a type (for example, starting with VS2002, MSVC has an __alignof() intrinsic). Those should be used when available.

Answer 4

As the others mentioned, its implementation dependant. Visual Studio 2005 uses 8 bytes as the default structure alignment. Internally, items are aligned by their size - a float has 4 byte alignment, a double uses 8, etc.
You can override the behavior with #pragma pack. GCC (and most compilers) have similar compiler options or pragmas.

Answer 5

It is possible to assume a structure alignment if you know more details about the compiler options that are in use. For example, #pragma pack(1) will force alignment on the byte level for some compilers.
Side note: I know the question was about alignment, but a side issue is padding. For embedded programming, binary data, and so forth -- In general, don't assume anything about structure alignment if possible. Rather use explicit padding if necessary in the structures. I've had cases where it was impossible to duplicate the exact alignment used in one compiler to a compiler on a different platform without adding padding elements. It had to do with the alignment of structures inside of structures, so adding padding elements fixed it.

Answer 6

If you want to find this out for a particular case in Windows, open up windbg:
Windbg.exe -z \path\to\somemodule.dll -y \path\to\symbols
Then, run:
dt somemodule!CSomeType

Answer 7

I don't think memory layout is guaranteed in any way in any C standard. This is very much vendor and architect-dependent. There might be ways to do it that work in 90% of cases, but they are not standard.
I would be very glad to be proven wrong, though =)

Answer 8

I agree mostly with Paul Betts, Ryan and Dan. Really, it's up to the developer, you can either keep the default alignment symanic's which Robert noted about (Robert's explanation is just the default behaviour and not by any means enforced or required), or you can setup whatever alignment you want /Zp[##].
What this means is that if you have a typedef with floats', long double's, uchar's etc... various assortments of arrays's included. Then have another type which has some of these oddly shaped members, and a single byte, then another odd member, it will simply be aligned at whatever preference the make/solution file defines.
As noted earlier, using windbg's dt command at runtime you can find out how the compiler laid out the structure in memory.
You can also use any pdb reading tool like dia2dump to extract this info from pdb's statically.

Answer 9

Modified from Peeter Joot's Blog
C structure alignment is based on the biggest size native type in the structure, at least generally (an exception is something like using a 64-bit integer on win32 where only 32-bit alignment is required).
If you have only chars and arrays of chars, once you add an int, that int will end up starting on a 4 byte boundary (with possible hidden padding before the int member). Additionally, if the structure isn’t a multiple of sizeof(int), hidden padding will be added at the end. Same thing for short and 64-bit types.
Example:
struct blah1 {
char x ;
char y[2] ;
};
sizeof(blah1) == 3
struct blah1plusShort {
char x ;
char y[2] ;
// <<< hidden one byte inserted by the compiler here
// <<< z will start on a 2 byte boundary (if beginning of struct is aligned).
short z ;
char w ;
// <<< hidden one byte tail pad inserted by the compiler.
// <<< the total struct size is a multiple of the biggest element.
// <<< This ensures alignment if used in an array.
};
sizeof(blah1plusShort) == 8

Answer 10

I read this answer after 8 years and I feel that the accepted answer from #Robert is generally right, but mathematically wrong.
To ensure alignment requirements for structure members, the alignment of a structure must be at least as strict as the least common multiple of the alignment of its members. Consider an odd example, where the alignment requirements of members are 4 and 10; in which case the alignment of the structure is LCM(4, 10) which is 20, and not 10. Of course, it is odd to see platforms with such alignment requirement which is not a power of 2, and thus for all practical cases, the structure alignment is equal to the maximum alignment of its members.
The reason for this is that, only if the address of the structure starts with the LCM of its member alignments, the alignment of all the members can be satisfied and the padding between the members and the end of the structure is independent of the start address.
Update: As pointed out by #chqrlie in the comment, C standard does not allow the odd values of the alignment. However this answer still proves why structure alignment is the maximum of its member alignments, just because the maximum happens to be the least common multiple, and thus the members are always aligned relative to the common multiple address.