Incorrect struct size

The size of the following struct should be 14 bytes. It appears to be 16 bytes. When I write it to a binary file, two bytes seem to be inserted after the first two bytes. The two erroneous bytes have the hex values: cc cc. The first four bytes of the struct when filled out appropriately are: 42 4D CC CC

This is the smallest amount of code that I could reproduce the problem

Code:

---

#include <iostream>

using std::cout;
using std::cin;

struct bfHeader
{
        unsigned short bfType;// = 19778 = 'BM'
        unsigned int bfSize;// = size of file
        unsigned short bfReserved1, bfReserved2;// = 0;
        unsigned int bfOffBits;// = 54;
};

int main()
{
        cout<<"\n Size of ushort: " << sizeof(unsigned short);
        cout<<"\n Size of uint: " << sizeof (unsigned int);
        cout<<"\n Expected size of struct: " << sizeof(unsigned short) * 3 + sizeof(unsigned int) * 2;
        cout<<"\n Size of bfHeader: " << sizeof(bfHeader);
        cin.get();
        return 0;
}

---

output

Code:

---

Size of ushort: 2
 Size of uint: 4
 Expected size of struct: 14
 Size of bfHeader: 16

---

This was done on VC++ 2008 express. Why is this struct 16 bytes in size?

For those that are curious, I discovered a page explaining what the compiler was doing

The size of my structure is not what I expect it to be

Compiler specific pragma fixes the problem

Code:

---

#pragma pack(push, 1)
struct bfHeader
{
        unsigned short bfType;// = 19778 = 'BM'
        unsigned int bfSize;// = size of file
        unsigned short bfReserved1, bfReserved2;// = 0;
        unsigned int bfOffBits;// = 54;
};
#pragma pack(pop)

---

structs are Dword ( 4 byte ) alligned by default

Aligning is not a very good idea unless you have a good reason to. Unaligned accesses can slow down your program and pragmas aren't portable.
Why do you need to? Why can you not just write it out using the >> operator and back using the << operator?

This is due to padding.

http://cboard.cprogramming.com/c-pro...sturcture.html

You should also arrange data types so that they slot together nicely:

Code:

---

struct s1
{
        int i;
        short j;
        int k;
        short l;
};

struct s2
{
        int i;
        int k;
        short j;
        short l;
};

---

The size of sizeof(s1) is 16 bytes and the sizeof(s2) is 12 bytes, because the two 2-byte short values can fit into a single 4-byte "slot".

It is likely the shorts will be padded, as well. Otherwise they will end up on unaligned addresses.

Quote:

---

Originally Posted by Ushakal

You should also arrange data types so that they slot together nicely:

---

The "slot" theory often doesn't work, as Elysia noted, because shorts are often aligned to a larger byte interval.

A common rule of thumb is to declare the members in order of decreasing size (i.e. big ones first). That doesn't guarantee no padding but is often a reasonable step towards minimising the total amount of padding added.

Since alignment of variables on (processor dependent) word boundaries often has a performance benefit most compilers add padding by default. For this reason, the sizes you quoted in your example are not guaranteed.