Question about byte alignment...

[chronozphere]

Hey all,

Consider the following record:

  TMyRecord = record
    ID:      Word;             //2 bytes
    recSize: Integer;          //4 bytes
    Str:     String;           //4 bytes
    b:       Boolean;          //1 bytes
    a:       char;             //2 bytes
  end;

In delphi 2010, when I enable {$A8}, sizeOf(TMyRecord) will return 16. Which strikes me because I thought each field would take 8bytes (or will be padded, until it takes 8 bytes).

It would correspond to the following memory layout:

WWxxxxxx
IIIIxxxx
SSSSxxxx
Bxxxxxxx
CCxxxxxx

However, this is not the case. How exactly are the fields packed together when I use 8-byte alignment?

Thanks

[virtual]

why not using packed record instead

[chronozphere]

why not using packed record instead?

I know about packed records, but it's not an answer to my question. I just want to know how exactly byte alignment works.

[VilleK]

I thought this was strange too, but I found this information here: http://www.delphibasics.co.uk/RTL.asp?Name=$Align&ExpandCode1=Yes

With $Align On (default), complex data types, such as records have their elements aligned to 2, 4 or 8 byte boundaries, as appropriate to the data type. For example, a Word field would be aligned to a 4 byte boundary.

Which means that A4 and A8 will only differ if the record contains 8 byte members such as Double or Int64.