Understanding alignment concept
c++, memory-alignment
Solution
Most CPU's have "preferences" about where data can be stored. When reading or writing to a memory address, the operation may be slower (or completely illegal) if the address doesn't match the data size you try to write. For example, it is common to require that 4-byte integers be allocated starting on an address that is divisible by 4.
That is, an `int` stored on address `7` is either less efficient, or completely illegal, depending on your CPU. But if it is stored at address `8`, the CPU is happy.
That is what alignment expresses: for any object of type `T` what must its address be divisible by, in order to satisfy the CPU's requirements?"
In C++, the alignment for an object is left implementation-defined (because, as said above, it depends on the CPU architecture). C++ merely says that every object has an alignment, and describes how to determine the alignment of compound objects.
Being "aligned for a `long double`" simply means that the object must be allocated so that its first byte is placed in an address that is valid for a `long double`. If the CPU architecture specifies the alignment of a `long double` to be 10 (for example), then it means that every object with this alignment must be allocated on an address that is divisible by 10.
Problem
An alignment is an implementation-defined integer value representing the number of bytes between successive addresses at which a given object can be allocated. That concept is a bit unclear. For instance: ``` struct B { long double d; }; struct D : virtual B { char c; } ``` When D is the type of a complete object, it will have a subobject of type B, so it must be aligned appropriately for a `long double`. What does it mean? `sizeof(long double)` is the number of bytes between what in that case??