struct size

[monarch_dodra]

Te reason it does that is because most computers perform much better if your memory is WORD aligned. Usually a WORD is 4 bytes, so your compiler will expand and ever rearrange your struct to make that happen. You can stop the compiler from doing any rearranging by wrapping it in extern "C"{}, but even that will cause padding for alignment.

This claim sparked my curiosity, as I was taught that elements in structs can't be re-arranged, ever, in C or C++. So I investigated a bit. I found these two interesting answers:

http://stackoverflow.com/questions/9.../916691#916691
http://stackoverflow.com/questions/1.../118177#118177

Nonstatic data members of a (non-union) class declared without an intervening access-specifier are allocated so that later members have higher addresses within a class object. The order of allocation of nonstatic data members separated by an access-specifier is unspecified (11.1)

Within a structure object, the non-bit-field members and the units in which bit-fields reside have addresses that increase in the order in which they are declared.

Long story short, it can't.

So yeah, that means the order of declaration can make a difference in the final struct size.

Unless you are on a very space-bound system though, I wouldn't worry too much about it, and declare elements in an order that makes sense to a human operator. And respect things like add new elements at the bottom, this makes it easier for change tacking.