Structures & sizeof operator

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In some cases, a structure has a member that wants the size of the structure. Is there a case where that member will be something other than the size of the structure.

If not, then does anyone know why the structure or what requires the structure doesn't do it itself?

Thanks,

Barry

Posted 24-Mar-11 8:46am

BarryPearlman

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Sergey Alexandrovich Kryukov · Accepted Answer · 2011-03-24T08:55:00

Yes, it can be different. Even though I consider it as a dirty style, this style is popular in Windows API.

Last field of structure is declared like SOME_TYPE[1], but the user of the structure is supposed to store the array of any other size in continues memory of the structure, at the very end. The API receives the pointer to the structure; so the only information about the actual size of the structure is passed in the Size field of the structure; usually this is a first field.

I guess this is done to avoid excessive pointers on pointers and dependency on memory allocation (when, say, the structure can be on stack but some memory for the fields is allocated in heap, so the user would need to clean up the structure after use).

—SA

CPallini · Accepted Answer · 2011-03-24T10:01:00

Often Windows API require struct size and of course there's a reason why.
I'll give a simple(r) example:

typedef struct TagMyData
{
   DWORD dwSize;
   DWORD dwVersion;
   BYTE bData[1];
} MyData;

int i;
MyData * pData = (MyData*) malloc(sizeof(MyData) + 9));
pData->dwVersion=0x00000001;
pData->dwSize = sizeof(MyData) + 9;
for (i=0;i<10;i++)
  pData->bData[i] = i;

/*
 ... 
 here you may call a function expecting such variable-size struct
 ...
*/

free(pData);

Here sizeof(MyData) is different with actual allocated memory.

User 7429338 · Accepted Answer · 2011-03-24T10:18:00

The size of the structure is fixed at compile time, this makes it possible to identify for which Windows version a program was designed, at execution time.

Suppose you use the OPENFILENAME[^] structure and you define your Windows version so that it will run on Windows NT 4.0. This means some members are not included in the structure when you compile it, and the sizeof() function will return a smaller result.
Now if you were to execute your program on Windows 2000 (NT 5.0), the lStructSize variable will make clear that the extra variables are not to be used because your program did not allocate memory for them.

So the conclusion is, this allows for expansion of the struct while maintaining backward compatibility.

Structures & sizeof operator

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