Why does a struct consisting of a char, short, and char (in that order), when compiled in C++ with 4-byte packing enabled, come to a 6-byte struct?

Question

I thought I understood how C/C++ handled struct member alignment. But I\'m getting strange results for a particular arrangement in Visual Studio 2008 and 2010.

Speci

Answer 1

From the MSDN documentation for #pragma pack (where n is the value you set):

The alignment of a member will be on a boundary that is either a multiple of n or a multiple of the size of the member, whichever is smaller.

sizeof(short) is two bytes, which is smaller than the packing value of four bytes that you set, so the short member is aligned to a two byte boundary.

The last char (c2) is padded with an extra byte after it so that when Alignment objects are placed in an array, the short element is still correctly aligned on a two-byte boundary. Array elements are contiguous and there can be no padding between them, so padding must be added to the end of the structure in order to ensure proper alignment in arrays.

Answer 2

Apparently, you indeed misunderstand it. In Visual Studio you cannot increase the alignment requirements for struct members of any type by using the struct packing settings. You can only decrease them.

If your struct consists of char (aligned at 1 byte boundary) and short (aligned at 2 byte boundary) objects, then using 4- and 8-byte packing settings will have absolutely no effect on the layout or size of your structure. The result will be exactly the same as with 2-byte packing. The structure will have size of 6 bytes.

The only packing setting that will have any effect in this case is 1-byte packing setting which will decrease the alignment requirement of short from 2 to 1 and result in 4 byte size of the structure.

Answer 3

The Visual Studio's C compiler with this command option /Zp[n] where a struct is packed on a n-byte boundary, this is where the #pragma pack directive comes in, members of the structure are aligned on a boundary that is multiple of n.