I have a struct with 2 members of type double and 1 member of type long.
The sizeof() function returns 24. Normaly it should be 20(2*8bytes for the doubles and 4 bytes for the long).
When I add a another member of type long sizeof() returns 32.
Another struct with 3 longs returns 12 with sizeof().
I suppose it has something to do with allignment and optimalisation? How can I avoid this.
Luc Vercruysse
A struct always gets its size adjusted to be n times the member with the highest alignment requirement.
If you have an 8-byte double in the struct, then the struct must be n*8 bytes large, so your first example will need four bytes of padding.
If you place the two long values directly after each other, then the modified struct should stay at 24 bytes. If you have one or more double between the two long, then the compiler needs two padding areas of four bytes each, resulting in a total size of 32 byte.
"on an ARM is it generally more efficient to leave the smallest structure elements for the end of the structure"
The general case for any processor that can take advantage of alignment is that you get the smallest and most efficient structures if the data is grouped on increasing or decreasing align requirements.
one long and one short and two byte can be grouped into 8 bytes if the align sizes are sorted in increasing or decreasing order. Any other combination will result in extra padding.
If you really need to get rid of the 'holes' in your structure, then you should consider 'packing' them.
Refer to:
www.keil.com/.../armccref_cjafjhjd.htm
Note that doing this can have a negative effect on code speed and size.
yes, because the compiler *must* assume each assess to the structure might be unaligned, which means extra shifts.
Thanks for very fast reply !!!