On Wed, Jun 14, 2017 at 9:09 PM, Luc Van Oostenryck
<luc.vanoostenryck@xxxxxxxxx> wrote:
> On Wed, Jun 14, 2017 at 10:25:59PM +0100, Dibyendu Majumdar wrote:
>> Hi,
>>
>> I am trying to understand what happens when multiple input files are
>> given to sparse. Are the symbols from previous files dropped when
>> sparse moves on to the next file?
>>
>> I see a call to remove_symbol_scope() in end_scope() implying that
>> symbols are dropped - but then I am not sure I am understanding this
>> correctly.
Yes, remove_symbol_scope() just remove symbol from the hash table
link list. After removal, the symbol will not be able to found by
lookup_symbol()
again.
>
> I never yet looked what's happening exactly but I also had seen
> a problem when using multiple files, at least a difference
> between sparse & gcc. The difference can quickly be seen
> with an input file like:
> int a = 1;
> int foo(void) {}
> and then call sparse or gcc with twice the same file, like
> 'sparse $file.c $file.c' or 'gcc -c $file.c $file.c'.
>
> gcc seems to process the files independently and it will simply
> overwrite the first $file.o when processing the second one and
> this without giving a warning about 'a' or 'foo()' being redefined.
>
> sparse seems to process the files more as if being concatenated
> (even if the file scope is respected) or like it would use
> a "super-scope" above the file scope of the input files and
> will complain that 'a' and 'foo()' are redefined.
The problem you are seeing is that, sparse remove the file
scope symbols. It does not remove extern symbols. It does have
a difference compare to gcc. I think the effect is that, currently
sparse does not have a linker, if you give more than one file to
sparse. Those file will be as if they are link together (more or less).
Chris
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