Re: [PATCH v2 11/19] gendwarfksyms: Limit structure expansion

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On 8/15/24 19:39, Sami Tolvanen wrote:
> Expand each structure type only once per exported symbol. This
> is necessary to support self-referential structures, which would
> otherwise result in infinite recursion, but is still sufficient for
> catching ABI changes.
> 
> For pointers to structure types, limit type expansion inside the
> pointer to two levels. This should be plenty for detecting ABI
> differences, but it stops us from pulling in half the kernel for
> types that contain pointers to large kernel data structures, like
> task_struct, for example.

I'm quite worried about this optimization for pointer types. It could
result in some kABI changes not being recognized.

I assume the goal of the optimization is to speed up the tool's runtime.
How much does it improve the processing time and is there any other way
how it could be done?

> diff --git a/scripts/gendwarfksyms/dwarf.c b/scripts/gendwarfksyms/dwarf.c
> index 92b6ca4c5c91..2f1601015c4e 100644
> --- a/scripts/gendwarfksyms/dwarf.c
> +++ b/scripts/gendwarfksyms/dwarf.c
> [...]
> @@ -651,6 +742,7 @@ static int process_exported_symbols(struct state *state, struct die *cache,
>  		else
>  			check(process_variable(state, &state->die));
>  
> +		cache_clear_expanded(&state->expansion_cache);
>  		return 0;
>  	default:
>  		return 0;

I wonder if it would make sense to share the cache between processing
individual exported symbols.

The hard case looks to be the following:
s#A struct A { int ; }
s#B struct B { s#A ; }
foo void foo ( s#B )
bar void bar ( s#A , s#B )

When processing foo, the code would cache s#B with expanded s#A.
However, when processing bar and reaching s#B, the cache should report
a miss because s#B with unexpanded s#A is required.

So the code would need to track which types were already expanded and
have each cache entry accordingly tagged with similar data.

Hm, it might be that doing all this additional tracking would then be
actually slower than processing the types repeatedly for each symbol.
I'm not sure.

-- 
Thanks,
Petr




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