allignment of VARiables in memory

janb

Hi,
I was expecting the memory allocated·in the VAR block matches seqence of definitions.

I'm puzzeld why

---

VAR
·· byte a[noparse][[/noparse]4]
·· word w1,w2
·· byte x···

---

results with
@x-@w2= 6

x·follows w2 and the difference in adresses should be equal to the size of word=2 ?

If I change dimention of a[noparse][[/noparse]4]-- >a[noparse][[/noparse]8] I'd get
@x-@w2= $A· which is 2+8

If I move decalartion of a[noparse]/noparse below declaration of x· I'd get 2 as difference - as expected.

The example code is attached.
Could you point where·my logic breaks?
Thanks
Jan

Phil Pilgrim (PhiPi)

I suspect that variables are allocated from large to small. So longs would be allocated first, then words, then bytes. This helps to minimize vacant bytes left between differently-aligned variables.

-Phil

janb

Hi,

I see.

In the fullDuplexSerial code there is a bunch of variables allocated in SPIN and the actual data transmision is done in assembler, assuming the order of variables as in SPIN VAR-block. Does it mean it works correctly only because all variables·were declared as long?

In other words, if I want to use some set of variables as a FORTRAN-style common block and pass only address of the first one I'd better decalre all of them as long to avoid surprises?

Thanks

Jan

Mike Green

Yes, you can assume that a block of same-size variables will be allocated in order, but not mixed size.· Local variables in a method are always longs so they are always allocated in order.· It would be nice to have this clearly stated in the documentation.· Maybe next iteration when there are "advanced" examples included.