Some basic concepts in FrictPhys.cpp

The source code website is:
https://gitlab.com/yade-dev/trunk/blob/master/pkg/dem/FrictPhys.cpp

Can i change the status of this question? I make a mistake to set it as solved. :(

Hello,

1.1

> Is the shared_ptr<FrictMat> a template class

yes. But as C++ newcomer, you can consider shared_ptr<SomeClass> as normal pointer SomeClass* with "automatic destruction"

> will instantiate a pointer to FricMat?

no, the pointer already exists. It just change (cast) it for C++ from
const shared_ptr<Material>&
to
const shared_ptr<FrictMat>&

> I have checked the definition of shared_ptr but it is too complicated for a c++ newer to understand.

apart from definitions, try to google some simple examples.

1.2

> The YADE_PTR_CAST will do change the b1 type to FrictMat?

yes, see above

> But if b1 is not a pointer to FrictMat, like a CohFrictMat, will it cause a material type mismatch between the particle material and the IPhysFunctor?

if b1 is not FrictMat, than mat1 will be nullptr, causing segmentation fault. But this function is called from Yade internals, specifically IPhysDispatcher, who sends there only correct types.

CohFrictMat is not a good example, because CohFrictMat **IS** FrictMat (it is derived from it) and the code would work OK.

2
> Why these two values could be negative?

who not? :-)
This condition is there from (at least, was too lazy to search more) 2009, probably its meaning was Sphere-Facet interaction or something like that

3
const shared_ptr<FrictPhys>& contactPhysics
means that the pointer is constant (the FrictPhys instance), not its content. You can change its content, but you cannot change, where the pointer points.

4
kn is MatchMaker [1]

cheers
Jan

[1] https://gitlab.com/yade-dev/trunk/commit/56597aa634e5f5a7684210dd4c8a858be5116cae#b34548f45a5a6c26d6a5293e702d6ccb1cbd5bbf_44_50
[2] https://yade-dem.org/doc/yade.wrapper.html#yade.wrapper.MatchMaker

>const shared_ptr<FrictMat>& mat1 = YADE_PTR_CAST<FrictMat>(b1);

This is saying that no matter what material class b1 actually has, mat1 is going to treat it as if it is a FrictMat. This means that we can use mat1 later to access members that FrictMat would have (e.g. mat1->young)

>But if b1 is not a pointer to FrictMat, like a CohFrictMat, will it cause a material type mismatch between the particle material and the IPhysFunctor?

Nope.

>shared_ptr but it is too complicated for a c++
[1]

> The Ra, Rb here are the stiffness related particle size? Why these two values could be negative?

No. https://yade-dem.org/doc/yade.wrapper.html#yade.wrapper.GenericSpheresContact.refR1

> Also the conversion of interaction->phys to FrictPhy type. Another is that since the reference contactPhysics has decoration of const, why it is able to change the value of its pointing memory

I think ti is because we are referencing using &. A bit confusing, no doubt.

>4. Real Kn = (!kn) ? 2*kna*knb/(kna+knb) : (*kn)(mat1->id,mat2->id,kna,knb);

kn is a shared_ptr<Matchmaker> with a custom () operator [2].

[1]https://www.yade-dem.org/doc/prog.html#shared-pointers

[2https://gitlab.com/yade-dev/trunk/blob/master/pkg/common/MatchMaker.hpp#L38

Ah we worked on this at the same time, Jan ;-).

About this piece:

const shared_ptr<FrictPhys>& contactPhysics

According to stack overflow, a const shared_ptr<FrictPhys> contactPhys (sans &), would mean that contactPhysics is constant [1]. So I am guessing we can only modify it because we are taking it as a reference too. Maybe you can clarify this?

[1]https://stackoverflow.com/questions/17793333/difference-between-const-shared-ptrt-and-shared-ptrconst-t

> According to stack overflow, a const shared_ptr<FrictPhys> contactPhys (sans &), would mean that contactPhysics is constant [1].

I don't think so. With 'const shared_ptr<FrictPhys> contactPhys ' the pointed data would not be constant, as opposed to 'shared_ptr<const FrictPhys> contactPhys '.

The role of '&' is only to avoid reference counting (it is the conventional method to pass shared_ptr).

Ah my mistake for p vs *p

Thanks very much for your answers. Helps me a lot.