[llvm-dev] How to efficiently extract the calledFunction from a complex CallInst?

[David Blaikie via llvm-dev]

On Thu, Nov 12, 2015 at 8:51 PM, Shen Liu <shl413 at lehigh.edu> wrote:

> Thanks for your helpful answer David! But one thing still makes me
> confused is when i use isa<CallInst> to check the original instruction the
> answser is yes.
> , which implies it is a callinst. Since it is not a CallInst, why does
> llvm add an opcode "call" before it? It looks a little weird here.
>

It is a CallInst, hence whwy you can call getCalledFunction - but you just
don't get an answer. It's a call, but not to a function. It's a call to a
bitcast of a function. So the operand(0) is not a Function, but a
BitCastInst (or a constant, I forget how they work).


>
> Shen
>
> On Thu, Nov 12, 2015 at 10:48 PM, David Blaikie <dblaikie at gmail.com>
> wrote:
>
>>
>>
>> On Thu, Nov 12, 2015 at 7:27 PM, Shen Liu via llvm-dev <
>> llvm-dev at lists.llvm.org> wrote:
>>
>>> Hi all,
>>>
>>> Usually if we want to get the called Function we can directly use
>>> CallInst->getCalledFunction(), however, today i encounter an unusual
>>> CallInst as follows:
>>>
>>>  %call11 = call double (...)* bitcast (double ()* @quantum_frand to
>>> double (...)*)()
>>>
>>> the original C source involve type cast:
>>>
>>> float u,v;
>>> extern double quantum_frand();
>>>    u = 2 * quantum_frand() - 1;
>>>    v = 2 * quantum_frand() - 1;
>>>
>>> In this case, CallInst->getCalledFunction() returns a nullptr unusually,
>>> I printed out the getOperand(0) and found the operand is the whole thing of
>>>  "double (...)* bitcast (double ()* @quantum_frand to double (...)*)()".
>>> Any member function calling on that fails so i don't know whether there is
>>> an efficient way to exactly get the called function @quantum_frand(...)
>>> here?
>>>
>>
>> What you have is a BitCastInst, so you'd have to downcast the Value* to
>> that, then look through the bitcast.
>>
>> But realize you may never actually find a function in the end:
>>
>> void func(void (*f)(void)) {
>>   f();
>> }
>>
>> the f call is a call to a function pointer, so there is no statically
>> knowable called function here. Your code will probably need to handle the
>> case where this arises.
>>
>> - David
>>
>>
>>> Thanks!
>>>
>>>
>>> Best regards,
>>>
>>> Shen
>>>
>>>
>>>
>>>
>>> _______________________________________________
>>> LLVM Developers mailing list
>>> llvm-dev at lists.llvm.org
>>> http://lists.llvm.org/cgi-bin/mailman/listinfo/llvm-dev
>>>
>>>
>>
>
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