Why doesn't size of struct is equals to sum of sizes of its individual member types?

Answer 1

It is because the compiler uses padding to bring each element into word alignment that is specific to the architecture for which you are compiling.

It can be for one of several reasons but usually:

1. Because some CPU's simply cannot read a long or long long multi-byte value when it isn't on an address multiple of its own size.

2. Because CPU's that can read off-aligned data may do it much slower than aligned.

You can often force it off with a compiler-specific directive or pragma.

When you do this, the compiler will generate relatively inefficient code to access the off-aligned data using multiple read/write operations.

Answer 2

The compiler can insert padding between members or at the end of the struct. Padding between members is typically done to keep the members aligned to maximize access speed. Padding at the end is done for roughly the same reason, in case you decide to create an array of the structs.

To prevent it from happening, you use something like #pragma pack(1).

Answer 3

This is called padding; which involves adding some more bytes in order to align the structure on addresses that are divisible by some special number, usually 2, 4, or 8. A compiler can even place padding between members to align the fields themselves on those boundaries.

This is a performance optimization: access to aligned memory addresses is faster, and some architectures don't even support accessing unaligned addresses.

For VC++, you can use the pack pragma to control padding between fields. However, note that different compilers have different ways of handling this, so if, for example, you also want to support GCC, you'll have to use a different declaration for that.