Re: [avr-gcc-list] avr-gcc generating ldd r31, Z+

[Jens Andersen]

Hi Ken

I have checked the Atmel instruction set document, and to me it
seems that the gcc generated code is ok.

As i read it, it says that if you load (Z) r30,r31 using the
Z+ (post-increment) or -Z (pre-decrement) form the result is undefined.

The gcc code uses the Z+q (displacement) form of the ldd instruction
which leaves Z unchanged, so it should be ok.

I hope this helps you.

Jens Andersen
 

ken restivo wrote:
> 
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> I seem to have found an avr-gcc/as bug, or perhaps I am doing something wrong 
> here.
> 
> Here is the relevant section of disassembly:
> 
>         t = t->next;
>  152:   07 80           ldd r0, Z+7 ; 0x07
>  154:   f0 85           ldd r31, Z+8    ; 0x08
>  156:   e0 2d           mov r30, r0
> 
> The databook states that ldd r31, Z+ is undefined, and simulavr is honoring 
> that and halting at instruction 0x154.
> 
> But t = t->next seems like a straightforward idiom. It works fine with gcc on 
> i386, of course, and even works in avr-gcc in a simple test program, if the X 
> and Z registers are free for use. But not here, though, for some reason I 
> don't understand.
> 
> At this point in the function's execution, most of the call-used registers 
> are in use, and a few of the call-saved registers too... there aren't a lot 
> of free registers floating around. So I could understand if there was simply 
> no place else to put this except to over-use the Z register. But why would 
> avr-gcc use the LDD instruction in a way that is undefined? I would guess 
> that it would move the items in the Z reg out of the way, save stuff onto the 
> stack, or at least fail to compile with an error?  I wonder if I am not doing 
> something that I shouldn't be?
> 
> Any help or clues would be appreciated.
> 
> - -ken
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