[llvm-dev] Function call replacement

[David Blaikie via llvm-dev]

On Wed, Mar 3, 2021 at 4:50 PM zxhuan <huanzhixuan at gmail.com> wrote:

> One example can be a constructor of a class, say class Foo, in C++. When
> you have a vector of Foo instances, the constructor will be called by the
> library function of vector.
>

Even in this example, you'd find the call to Foo's constructor in the
module somewhere. The "library function of vector" would be instantiated in
this same translation unit and emitted into the LLVM IR module.

If the call is really coming from outside the module there's not much you
can do to effect it.


> There are many such functions, that you can't know they are actually
> executed until the runtime.
>
> Beside that, I have done some research today. This has to be done by
> modifying the class's vtable.
>

If the function is virtual, and if you have access to the vtable (perhaps
it's in another module/translation unit too).


> I have manually modified the IR file and it works. But I just can't find
> an API function of LLVM to modify the vtable of the class.
>

LLVM IR doesn't know much about vtables - it's "just" another array of
function pointers as far as LLVM is concerned. But you could modify that
array if you like.


>
> On Wed, Mar 3, 2021 at 3:03 PM David Blaikie <dblaikie at gmail.com> wrote:
>
>>
>>
>> On Wed, Mar 3, 2021 at 11:55 AM zxhuan via llvm-dev <
>> llvm-dev at lists.llvm.org> wrote:
>>
>>> Hi all,
>>>
>>> I am trying to replace calls to a function, say foo() with calls to
>>> my_foo(). In the IR file, there is no call to foo(). However, foo() will be
>>> called during execution.
>>>
>>
>> Can't say I'm following here ^ there is no call to foo() but it is called
>> during execution? Is it called indirectly (eg: void (*x)() = foo; x(); )?
>> is the call in another module/not this IR file?
>>
>>
>>> I can't use the conventional way to do it (i.e. locate the call
>>> instruction, create a call site and then create a call to my_foo(), then
>>> replace all uses with the return value of my_foo()), because you can't
>>> locate the call instruction. So I tried to swap the name of foo() and
>>> my_foo(). However, this only results in the original function with the new
>>> name being called. I know I must have missed some steps, resulting in the
>>> reference to the original function still existing after the renaming. The
>>> question is, what is the missing step? Thank you so much for reading my
>>> description. I hope I make my question clear.
>>>
>>> Best,
>>> Jason
>>> _______________________________________________
>>> LLVM Developers mailing list
>>> llvm-dev at lists.llvm.org
>>> https://lists.llvm.org/cgi-bin/mailman/listinfo/llvm-dev
>>>
>>
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