I'm planning to use std::string as a generic data buffer (instead of roll my own). I need to pack all kinds of POD into it, including user defined structs, is the memory buffer allocated by std::string properly aligned for such purpose ?
If it's unspecified in C++ standard, what's the situation in libstdc++ ?
The host CPU is x86_64.
First of all, std::string is probably not the best container to use if what you want to do is store arbitrary data. I'd suggest using std::vector instead.
Second, the alignment of all allocations made by the container is controlled by its allocator (the second template parameter, which defaults to std::allocator<T>). The default allocator will align allocations on the size of the largest standard type, which is often long long or long double, respectively 8 and 16 bytes on my machine, but the size of these types is not mandated by the standard.
If you want a specific alignment you should either check what your compiler aligns on, or ask for alignment explicitly, by supplying your own allocator or using std::aligned_storage.
Related
I'm running tests to see how variables are getting placed on the memory and sizing them out when I use a struct.
Consider I have a struct that looks like below:
typedef struct _ttmp
{
WCHAR wcsTest1[13];
WCHAR wcsTest2[13];
wstring wstr;
}TTMP, *LPTTMP;
How big is the size of TTMP when STL classes like wstring are dynamically allocated?
Am I treating wstr as a 4-byte pointer?
I ran some tests to see the size of TTMP and got the size of the struct to be 88-bytes
and two of WCHAR arrays were 26-bytes which leaves the size of wstr to be 36-bytes, but that 36-bytes does not really make sense if I were to treat the wstring as a pointer. Seems like alignment padding does not apply here since I'm only using 32-bit variables.
Also, would it be bad a practice to use ZeroMemory api on structs with STL?
I've heard from someone that it is not safe to use the api, but the program ran fine when I test it
sizeof(WCHAR[13])=26, which is not an even multiple of 4 or 8, so alignment padding would account for a few bytes (unless you set the struct's alignment to 1-2 bytes via #pragma pack(1) or #pragma pack(2) or equivalent).
But, std::(w)string is much more than just a single pointer. It may have a few pointers in it, or at least a pointer and a couple of integers, which it uses to determine its c_str()/data(), size() and capacity() values. And it may even have an internal fixed buffer for use in Small String Optimization, which avoids dynamic memory allocation of small string values (this is likely true in your situation). So, at a minimum, a std::(w)string instance could be as few as, say, 8 bytes, or it could be as many as, say, 40 bytes, depending on its internal implementation.
See Exploring std::string for more details.
In your case, sizeof(std::wstring) is likely 32 (that is what gcc uses, for instance). So, 88-26-26-32=4, and those 4 bytes could easily be accounted for by alignment padding.
And yes, ZeroMemory() would be very bad (ie, undefined behavior) to use on a struct containing non-POD members.
I need a data structure in C++ that acts like a standard container of bytes but aligns the buffer at a multiple of four bytes. I'd like to re-use standard library abstractions as much as possible, rather than rolling my own abstraction.
Until now, I had been using std::string and std::vector<std::uint8_t> for this purpose. Unfortunately, I've gotten bug reports on the latest Mac OS, where apparently string::data() is no longer 4-byte aligned, but rather rather at an address congruent to 1 mod 4. As soon as I saw this, I realized of course nothing in the spec guarantees strings will be 4-byte aligned. I could switch over to vector<char>, but unfortunately now I'm not sure why this should be 4-byte aligned. Potentially even with a custom allocator the vector implementation could do something strange at the beginning of the buffer it allocates.
My question: What is a simple way of getting a dynamically-sized container of single-byte objects from the C++ standard library in which the first byte is at a 4-byte aligned address and individual bytes can be accessed through operator[]?
Note that this is not the same thing as asking how to ensure that the allocator used by the container returns 4-byte aligned memory. For example, std::string still allocates 4-byte aligned memory (probably 8, actually), it's just that on Mac OS string::data() does not point to the start of the allocated buffer. I don't see anything in the spec that would prevent a vector<char> from doing the same thing, even though for now that seems to work.
One of the solution is to use std::vector<uint32_t> internally, encapsulate that and convert data() to unsigned char * when you use it.
Usually data is aligned at power of two addresses depending on its size.
How should I align a struct or class with size of 20 bytes or another non-power-of-two size?
I'm creating a custom stack allocator so I guess that the compiler wont align data for me since I'm working with a continuous block of memory.
Some more context:
I have an Allocator class that uses malloc() to allocate a large amount of data.
Then I use void* allocate(U32 size_of_object) method to return the pointer that where I can store whether objects I need to store.
This way all objects are stored in the same region of memory and it will hopefully fit in the cache reducing cache misses.
C++11 has the alignof operator specifically for this purpose. Don't use any of the tricks mentioned in other posts, as they all have edge cases or may fail for certain compiler optimisations. The alignof operator is implemented by the compiler and knows the exact alignment being used.
See this description of c++11's new alignof operator
Although the compiler (or interpreter) normally allocates individual data items on aligned boundaries, data structures often have members with different alignment requirements. To maintain proper alignment the translator normally inserts additional unnamed data members so that each member is properly aligned. In addition the data structure as a whole may be padded with a final unnamed member. This allows each member of an array of structures to be properly aligned. http://en.wikipedia.org/wiki/Data_structure_alignment#Typical_alignment_of_C_structs_on_x86
This says that the compiler takes care of it for you, 99.9% of the time. As for how to force an object to align a specific way, that is compiler specific, and only works in certain circumstances.
MSVC: http://msdn.microsoft.com/en-us/library/83ythb65.aspx
__declspec(align(20))
struct S{ int a, b, c, d; };
//must be less than or equal to 20 bytes
GCC: http://gcc.gnu.org/onlinedocs/gcc-3.4.0/gcc/Type-Attributes.html
struct S{ int a, b, c, d; }
__attribute__ ((aligned (20)));
I don't know of a cross-platform way (including macros!) to do this, but there's probably neat macro somewhere.
Unless you want to access memory directly, or squeeze maximum data in a block of memory you don't worry about alignment -- the compiler takes case of that for you.
Due to the way processor data buses work, what you want to avoid is 'mis-aligned' access. Usually you can read a 32 bit value in a single access from addresses which are multiples of four; if you try to read it from an address that's not such a multiple, the CPU may have to grab it in two or more pieces. So if you're really worrying about things at this level of detail, what you need to be concerned about is not so much the overall struct, as the pieces within it. You'll find that compilers will frequently pad out structures with dummy bytes to ensure aligned access, unless you specifically force them not to with a pragma.
Since you've now added that you actually want to write your own allocator, the answer is straight-forward: Simply ensure that your allocator returns a pointer whose value is a multiple of the requested size. The object's size itself will already come suitably adjusted (via internal padding) so that all member objects themselves are properly aligned, so if you request sizeof(T) bytes, all your allocator needs to do is to return a pointer whose value is divisible by sizeof(T).
If your object does indeed have size 20 (as reported by sizeof), then you have nothing further to worry about. (On a 64-bit platform, the object would probably be padded to 24 bytes.)
Update: In fact, as I only now came to realize, strictly speaking you only need to ensure that the pointer is aligned, recursively, for the largest member of your type. That may be more efficient, but aligning to the size of the entire type is definitely not getting it wrong.
How should I align a struct or class with size of 20 bytes or another non-power-of-two size?
Alignment is CPU-specific, so there is no answer to this question without, at least, knowing the target CPU.
Generally speaking, alignment isn't something that you have to worry about; your compiler will have the rules implemented for you. It does come up once in a while, like when writing an allocator. The classic solution is discussed in The C Programming Language (K&R): use the worst possible alignment. malloc does this, although it's phrased as, "the pointer returned if the allocation succeeds shall be suitably aligned so that it may be assigned to a pointer to any type of object."
The way to do that is to use a union (the elements of a union are all allocated at the union's base address, and the union must therefore be aligned in such a way that each element could exist at that address; i.e., the union's alignment will be the same as the alignment of the element with the strictest rules):
typedef Align long;
union header {
// the inner struct has the important bookeeping info
struct {
unsigned size;
header* next;
} s;
// the align member only exists to make sure header_t's are always allocated
// using the alignment of a long, which is probably the worst alignment
// for the target architecture ("worst" == "strictest," something that meets
// the worst alignment will also meet all better alignment requirements)
Align align;
};
Memory is allocated by creating an array (using somthing like sbrk()) of headers large enough to satisfy the request, plus one additional header element that actually contains the bookkeeping information. If the array is called arry, the bookkeeping information is at arry[0], while the pointer returned points at arry[1] (the next element is meant for walking the free list).
This works, but can lead to wasted space ("In Sun's HotSpot JVM, object storage is aligned to the nearest 64-bit boundary"). I'm aware of a better approach that tries to get a type-specific alignment instead of "the alignment that will work for anything."
Compilers also often have compiler-specific commands. They aren't standard, and they require that you know the correct alignment requirements for the types in question. I would avoid them.
I'm hoping that the elements are 1 byte aligned and similarly that a std::vector<int> is 4 byte aligned ( or whatever size int happens to be on a particular platform ).
Does anyone know how standard library containers get aligned?
The elements of the container have at least the alignment required for them in that implementation: if int is 4-aligned in your implementation, then each element of a vector<int> is an int and therefore is 4-aligned. I say "if" because there's a difference between size and alignment requirements - just because int has size 4 doesn't necessarily mean that it must be 4-aligned, as far as the standard is concerned. It's very common, though, since int is usually the word size of the machine, and most machines have advantages for memory access on word boundaries. So it makes sense to align int even if it's not strictly necessary. On x86, for example, you can do unaligned word-sized memory access, but it's slower than aligned. On ARM unaligned word operations are not allowed, and typically crash.
vector guarantees contiguous storage, so there won't be any "padding" in between the first and second element of a vector<char>, if that's what you're concerned about. The specific requirement for std::vector is that for 0 < n < vec.size(), &vec[n] == &vec[0] + n.
[Edit: this bit is now irrelevant, the questioner has disambiguated: The container itself will usually have whatever alignment is required for a pointer, regardless of what the value_type is. That's because the vector itself would not normally incorporate any elements, but will have a pointer to some dynamically-allocated memory with the elements in that. This isn't explicitly required, but it's a predictable implementation detail.]
Every object in C++ is 1-aligned, the only things that aren't are bitfields, and the elements of the borderline-crazy special case that is vector<bool>. So you can rest assured that your hope for std::vector<char> is well-founded. Both the vector and its first element will probably also be 4-aligned ;-)
As for how they get aligned - the same way anything in C++ gets aligned. When memory is allocated from the heap, it is required to be aligned sufficiently for any object that can fit into the allocation. When objects are placed on the stack, the compiler is responsible for designing the stack layout. The calling convention will specify the alignment of the stack pointer on function entry, then the compiler knows the size and alignment requirement of each object it lays down, so it knows whether the stack needs any padding to bring the next object to the correct alignment.
I'm hoping that the elements are 1 byte aligned and similarly that a std::vector is 4 byte aligned ( or whatever size int happens to be on a particular platform ).
To put it simply, std::vector is a wrapper for a C array. The elements of the vector are aligned as if they were in the array: elements are guaranteed to occupy continues memory block without any added gaps/etc, so that std::vector<N> v can be accessed as a C array using the &v[0]. (Why vector has to reallocate storage sometimes when elements are added to it.)
Does anyone know how standard library containers get aligned?
Alignment of elements is platform specific but generally a simple variable is aligned so that its address is divisible by its size (natural alignment). Structures/etc are padded (empty filler space at the end) on the largest data type they contain to ensure that if the structure is put into an array, all fields would retain their natural alignment.
For other containers (like std::list or std::map) use data via template mechanics are made a part of internal structure and the structure is allocated by operator new. The new is guaranteed (custom implementation must obey the rule too; inherited from the malloc()) to return memory block which is aligned on largest available primitive data type (*). That is to ensure that regardless what structure or variable would be places in the memory block, it will be accessed in aligned fashion. Unlike std::vector, obviously, the elements of most other STL containers are not guaranteed to be within the same continuous memory block: they are newed one by one, not with new[].
(*) As per C++ standard, "The allocation function (basic.stc.dynamic.allocation) called by a new-expression (expr.new) to allocate size bytes of storage suitably aligned to represent any object of that size." That is a softer requirement compared to one malloc() generally abides, as per POSIX: "The pointer returned if the allocation succeeds shall be suitably aligned so that it may be assigned to a pointer to any type of object [...]". C++ requirement in a way reenforces the natural alignment requirement: dynamically allocated char would be aligned as char requires, but not more.
Do you mean the vector members, or the vector structure itself? Members are guaranteed to be contiguous in memory but structure alignment is platform/compiler-dependent. On Windows this can be set at compile-time and also overridden using #pragma pack().
The answer for other containers is likely not the same as for vector, so I would ask specific questions about the ones you care about.
The alignment of the whole container is implementation dependent. It is usually at least sizeof(void*), that is 4 or 8 bytes depending on the platform, but may be larger.
If special (guaranteed) alignment is needed use plain arrays or write/adapt some generic array class with:
// allocation
char* pointer = _mm_malloc(size, alignment);
// deallocation
_mm_free(pointer);
Is there any portable way to determine what the maximum possible alignment for any type is?
For example on x86, SSE instructions require 16-byte alignment, but as far as I'm aware, no instructions require more than that, so any type can be safely stored into a 16-byte aligned buffer.
I need to create a buffer (such as a char array) where I can write objects of arbitrary types, and so I need to be able to rely on the beginning of the buffer to be aligned.
If all else fails, I know that allocating a char array with new is guaranteed to have maximum alignment, but with the TR1/C++0x templates alignment_of and aligned_storage, I am wondering if it would be possible to create the buffer in-place in my buffer class, rather than requiring the extra pointer indirection of a dynamically allocated array.
Ideas?
I realize there are plenty of options for determining the max alignment for a bounded set of types: A union, or just alignment_of from TR1, but my problem is that the set of types is unbounded. I don't know in advance which objects must be stored into the buffer.
In C++11 std::max_align_t defined in header cstddef is a POD type whose alignment requirement is at least as strict (as large) as that of every scalar type.
Using the new alignof operator it would be as simple as alignof(std::max_align_t)
In C++0x, the Align template parameter of std::aligned_storage<Len, Align> has a default argument of "default-alignment," which is defined as (N3225 ยง20.7.6.6 Table 56):
The value of default-alignment shall be the most stringent alignment requirement for any C++ object type whose size is no greater than Len.
It isn't clear whether SSE types would be considered "C++ object types."
The default argument wasn't part of the TR1 aligned_storage; it was added for C++0x.
Unfortunately ensuring max alignment is a lot tougher than it should be, and there are no guaranteed solutions AFAIK. From the GotW blog (Fast Pimpl article):
union max_align {
short dummy0;
long dummy1;
double dummy2;
long double dummy3;
void* dummy4;
/*...and pointers to functions, pointers to
member functions, pointers to member data,
pointers to classes, eye of newt, ...*/
};
union {
max_align m;
char x_[sizeofx];
};
This isn't guaranteed to be fully
portable, but in practice it's close
enough because there are few or no
systems on which this won't work as
expected.
That's about the closest 'hack' I know for this.
There is another approach that I've used personally for super fast allocation. Note that it is evil, but I work in raytracing fields where speed is one of the greatest measures of quality and we profile code on a daily basis. It involves using a heap allocator with pre-allocated memory that works like the local stack (just increments a pointer on allocation and decrements one on deallocation).
I use it for Pimpls particularly. However, just having the allocator is not enough; for such an allocator to work, we have to assume that memory for a class, Foo, is allocated in a constructor, the same memory is likewise deallocated only in the destructor, and that Foo itself is created on the stack. To make it safe, I needed a function to see if the 'this' pointer of a class is on the local stack to determine if we can use our super fast heap-based stack allocator. For that we had to research OS-specific solutions: I used TIBs and TEBs for Win32/Win64, and my co-workers found solutions for Linux and Mac OS X.
The result, after a week of researching OS-specific methods to detect stack range, alignment requirements, and doing a lot of testing and profiling, was an allocator that could allocate memory in 4 clock cycles according to our tick counter benchmarks as opposed to about 400 cycles for malloc/operator new (our test involved thread contention so malloc is likely to be a bit faster than this in single-threaded cases, perhaps a couple of hundred cycles). We added a per-thread heap stack and detected which thread was being used which increased the time to about 12 cycles, though the client can keep track of the thread allocator to get the 4 cycle allocations. It wiped out memory allocation based hotspots off the map.
While you don't have to go through all that trouble, writing a fast allocator might be easier and more generally applicable (ex: allowing the amount of memory to allocate/deallocate to be determined at runtime) than something like max_align here. max_align is easy enough to use, but if you're after speed for memory allocations (and assuming you've already profiled your code and found hotspots in malloc/free/operator new/delete with major contributors being in code you have control over), writing your own allocator can really make the difference.
Short of some maximally_aligned_t type that all compilers promised faithfully to support for all architectures everywhere, I don't see how this could be solved at compile time. As you say, the set of potential types is unbounded. Is the extra pointer indirection really that big a deal?
Allocating aligned memory is trickier than it looks - see for example Implementation of aligned memory allocation
This is what I'm using. In addition to this, if you're allocating memory then a new()'d array of char with length greater than or equal to max_alignment will be aligned to max_alignment so you can then use indexes into that array to get aligned addresses.
enum {
max_alignment = boost::mpl::deref<
boost::mpl::max_element<
boost::mpl::vector<
boost::mpl::int_<boost::alignment_of<signed char>::value>::type,
boost::mpl::int_<boost::alignment_of<short int>::value>::type,
boost::mpl::int_<boost::alignment_of<int>::value>::type, boost::mpl::int_<boost::alignment_of<long int>::value>::type,
boost::mpl::int_<boost::alignment_of<float>::value>::type,
boost::mpl::int_<boost::alignment_of<double>::value>::type,
boost::mpl::int_<boost::alignment_of<long double>::value>::type,
boost::mpl::int_<boost::alignment_of<void*>::value>::type
>::type
>::type
>::type::value
};
}