[solved] Type conversion problem: TechDraw::BaseGeom* versus TechDraw::BaseGeomPtr

[edi]

Hi everybody,
I have the two functions:

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void BaseGeom::intersectionLL(TechDraw::BaseGeomPtr geom1,  TechDraw::BaseGeomPtr geom2, std::vector<Base::Vector3d>& interPoints)
{
	...
}


std::vector<Base::Vector3d> BaseGeom::intersection(TechDraw::BaseGeomPtr geom2)
{
	...
	std::vector<Base::Vector3d> interPoints;
	intersectionLL(this, geom2, interPoints); // <-- compiler message
}

...
TechDraw::BaseGeomPtr geom1 = ...
TechDraw::BaseGeomPtr geom2 = ...
interPoints = geom1->intersection(geom2);

The compiler returns the message:

error: cannot convert ‘TechDraw::BaseGeom*’ to ‘TechDraw::BaseGeomPtr’ {aka ‘std::shared_ptr<TechDraw::BaseGeom>’}

The compiler message occures, because "this" is of type TechDraw::BaseGeom*, it should be TechDraw::BaseGeomPtr.
How can I convert "this" ?

[wmayer]

You have to make BaseGeom a subclass of std::enable_shared_from_this. Then inside your function you can use shared_from_this().

Currently BaseGeom is declared in Geometry.h with

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class TechDrawExport BaseGeom
...

and you have to replace this line with

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class TechDrawExport BaseGeom : public std::enable_shared_from_this<BaseGeom>
...

[edi]

Thank you for the very quick answer. It was the solution.