llvm dev - Oct 2017 - Why does LLVm 3.8.0 recognize fputs_unlocked as a vararg function?

Considering F represents the function fputs_unlocked() in an LLVM pass,

 => F->isVarArg() returns true
 => F->getNumParams() returns 0
 => *F returns declare i32 @fputs_unlocked(...)

The signature of fputs_unlocked from man page is:
int fputs_unlocked(const char *s, FILE *stream);

Can anybody explain why fputs_unlocked() is recognized as a vararg method
while it accepts two fixed parameters?

-- 

Thanks & Regards,
Dipanjan
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On 27 October 2017 at 11:01, Dipanjan Das via llvm-dev
<llvm-dev at lists.llvm.org> wrote:
> Can anybody explain why fputs_unlocked() is recognized as a vararg method
> while it accepts two fixed parameters?
Are you sure the correct header has been included? Without that
declaration the legacy C rules dictate essentially that prototype.
Clang ought to give a warning about it though.

Tim.

I tested on a sample program and it seems to work correctly. Precisely, the
issue happens on base64/uniq/md5sum source code in LAVA-M (
http://moyix.blogspot.com/2016/10/the-lava-synthetic-bug-corpora.html) dataset
while instrumenting by DataFlowSanitizer pass of LLVM 3.8.0.


On 27 October 2017 at 03:22, Tim Northover <t.p.northover at gmail.com>
wrote:
> On 27 October 2017 at 11:01, Dipanjan Das via llvm-dev
> <llvm-dev at lists.llvm.org> wrote:
> > Can anybody explain why fputs_unlocked() is recognized as a vararg
method
> > while it accepts two fixed parameters?
>
> Are you sure the correct header has been included? Without that
> declaration the legacy C rules dictate essentially that prototype.
> Clang ought to give a warning about it though.
>
> Tim.
>


-- 

Thanks & Regards,
Dipanjan
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