hvladDimitry SibiryakovБесполезно. В Дельфи нет опции для "оптимального" выравнивания структур, которое
используется в Си и, соответственно, буферах.

D6, XE3Record Types When a record type is declared in the {$A+} state (the default), and when the declaration does not include a packed modifier, the type is an unpacked record type, and the fields of the record are aligned for efficient access by the CPU, and according to the platform. The alignment is controlled by the type of each field. Every data type has an inherent alignment, which is automatically computed by the compiler. The alignment can be 1, 2, 4, or 8, and represents the byte boundary on which a value of the type must be stored in order to provide the most efficient access. The table below lists the alignments for all data types.
Но
D6, XE3In the {$A8} or {$A+} state, fields in record types that are declared without the packed modifier and fields in class structures are aligned on quad-word boundariesПочему-то я первой цитате (как и здравому смыслу) верю больше.

PS Не проверял.
Вероятно, второе утверждение для 64-бит компиляции. В любом случае надо проверять структуры ABI для 32- и 64- отдельно.

/* troll
И зачем в учебном языке какое то выравнивание структур =)
*/