I am working on refactoring some old code and have found few structs containing zero length arrays (below). Warnings depressed by pragma, of course, but I've failed to create by "new" structures containing such structures (error 2233). Array 'byData' used as pointer, but why not to use pointer instead? or array of length 1? And of course, no comments were added to make me enjoy the process... Any causes to use such thing? Any advice in refactoring those?
struct someData
{
int nData;
BYTE byData[0];
}
NB It's C++, Windows XP, VS 2003
An array of ZERO length can be defined in JAVA. It can be declared in the following ways:- > int arrInt = new int[0];> int arrInt = new Int[]{};But this array will be unchangable.
It will just point to an array of size 0, which is perfectly valid.
This array is immutable (it can't be changed), and can be shared throughout the application.
In the case of a zero sized array, no objects are created, and no address is required to be given, hence it falls (in my opinion) outside of the standard's description of an array, becoming a valid extension of the language. In other words, a zero sized array is not an array, but something else (an extension).
Yes this is a C-Hack.
To create an array of any length:
struct someData* mallocSomeData(int size) { struct someData* result = (struct someData*)malloc(sizeof(struct someData) + size * sizeof(BYTE)); if (result) { result->nData = size; } return result; }
Now you have an object of someData with an array of a specified length.
There are, unfortunately, several reasons why you would declare a zero length array at the end of a structure. It essentially gives you the ability to have a variable length structure returned from an API.
Raymond Chen did an excellent blog post on the subject. I suggest you take a look at this post because it likely contains the answer you want.
Note in his post, it deals with arrays of size 1 instead of 0. This is the case because zero length arrays are a more recent entry into the standards. His post should still apply to your problem.
http://blogs.msdn.com/oldnewthing/archive/2004/08/26/220873.aspx
EDIT
Note: Even though Raymond's post says 0 length arrays are legal in C99 they are in fact still not legal in C99. Instead of a 0 length array here you should be using a length 1 array
This is an old C hack to allow a flexible sized arrays.
In C99 standard this is not neccessary as it supports the arr[] syntax.
Your intution about "why not use an array of size 1" is spot on.
The code is doing the "C struct hack" wrong, because declarations of zero length arrays are a constraint violation. This means that a compiler can reject your hack right off the bat at compile time with a diagnostic message that stops the translation.
If we want to perpetrate a hack, we must sneak it past the compiler.
The right way to do the "C struct hack" (which is compatible with C dialects going back to 1989 ANSI C, and probably much earlier) is to use a perfectly valid array of size 1:
struct someData
{
int nData;
unsigned char byData[1];
}
Moreover, instead of sizeof struct someData
, the size of the part before byData
is calculated using:
offsetof(struct someData, byData);
To allocate a struct someData
with space for 42 bytes in byData
, we would then use:
struct someData *psd = (struct someData *) malloc(offsetof(struct someData, byData) + 42);
Note that this offsetof
calculation is in fact the correct calculation even in the case of the array size being zero. You see, sizeof
the whole structure can include padding. For instance, if we have something like this:
struct hack {
unsigned long ul;
char c;
char foo[0]; /* assuming our compiler accepts this nonsense */
};
The size of struct hack
is quite possibly padded for alignment because of the ul
member. If unsigned long
is four bytes wide, then quite possibly sizeof (struct hack)
is 8, whereas offsetof(struct hack, foo)
is almost certainly 5. The offsetof
method is the way to get the accurate size of the preceding part of the struct just before the array.
So that would be the way to refactor the code: make it conform to the classic, highly portable struct hack.
Why not use a pointer? Because a pointer occupies extra space and has to be initialized.
There are other good reasons not to use a pointer, namely that a pointer requires an address space in order to be meaningful. The struct hack is externalizeable: that is to say, there are situations in which such a layout conforms to external storage such as areas of files, packets or shared memory, in which you do not want pointers because they are not meaningful.
Several years ago, I used the struct hack in a shared memory message passing interface between kernel and user space. I didn't want pointers there, because they would have been meaningful only to the original address space of the process generating a message. The kernel part of the software had a view to the memory using its own mapping at a different address, and so everything was based on offset calculations.
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