问题
I want to do something like this:
template <typename T:public Vertex> addTri( T v1, T v2, T v3 )
{
// Take v1.pos, v2.pos, v3.pos and create a geometric repn..
Triangle tri( v1.pos, v2.pos, v3.pos ) ; // all vertices will
// have to have a .pos member.
// Create the vertex buffer..
VertexBuffer<T> vb ...
}
Since that doesn't work, this is my workaround..
template <typename T> addTri( T v1, T v2, T v3 )
{
Vertex* p1 = (Vertex*)&v1 ;
// This is a very "shut up C++, I know what I'm doing" type cast.
// I'd like for C++ to know that all vertex types (T in this case)
// __will__ have a Vector member .pos.
Triangle tri( p1->pos, p2->pos, p3->pos ) ;
// Create the vertex buffer..
VertexBuffer<T> vb ...
}
Background
In case you're interested, I'm trying to write a general bit of code to handle triangle creation.
Each vertex has to have a .pos member, because each vertex has to have a position in space.
However not every vertex type will have a texture coordinate. Not every vertex will have a color. Hence the parameterized types.
A similar approach is used in XNA VertexBuffer.SetData<T>.
回答1:
You cannot specify a type restriction in the template type argument. However, generally, you don't have to.
If you simply do:
template <typename T> addTri( T v1, T v2, T v3 )
{
Vertex &v1r = v1;
// ....
}
This will work if the function is instantiated with a derivative of Vertex. It will create an (obscure) error if T & is not convertible to Vertex &.
If you don't even care if the types are convertible to Vertex as long as they have the same members, you can even skip the assignment - C++ template arguments essentially work using duck typing; if you do v1.x, and T contains a member named x, then it will work, whatever type T might actually be.
You can be a bit more sophisticated using boost's type-traits library and a static assertion; with this, you can start defining an assertion to make the error a bit easier to understand:
template <typename T> addTri( T v1, T v2, T v3 )
{
BOOST_STATIC_ASSERT_MSG(boost::is_convertible<T&, Vertex&>::value,
"Template argument must be a subclass of Vertex");
Vertex &v1r = v1;
// ....
}
回答2:
A combination of enable_if, is_base_of and is_convertible typetraits should do the job:
template <typename T>
struct Foo : public std::enable_if<std::is_base_of<YourBase, T>::value &&
std::is_convertible<T&, A&>::value,
T>::type
{
// consider "using YourBase::foo;" directives here
};
The type traits are available from <type_traits> in modern compilers, or <tr1/type_traits> or Boost otherwise.
回答3:
You can do:
#include <type_traits>
template <typename T>
void addTri(T v1, T v2, T v3, char (*)[is_base_of<Vertex, T>::value] = 0)
{
...
}
to disable the generation of addTri if T doesn't inherit from Vertex. But you don't need it to be able to use the pos member.
Update: Actually std::is_base_of will return true if Vertex is an inaccessible base class of T. Use the following implementation of is_base_of instead:
template <typename B, typename D>
struct is_base_of
{
static const bool value = std::is_convertible<D*, B*>::value
&& std::is_class<B>::value;
};
回答4:
Just use your first solution without the odd :public Vertex. When you instantiate it with a Vertex or with something that just has a pos member, it will be fine. C++ doesn't have to know that every T has a pos member. If you anytime instantiate the template with anything that has no pos member, you will get a compiler error, otherwise it's fine.
What you're looking for is concepts, but they have been dropped from the C++0x standard, I think.
回答5:
You may be looking at templates wrong. What you describe looks to be handled better by good ole-fashioned inheritance. Instead of passing instances of your object, try passing pointers instead like this:
addTri( Vertex *v1, Vertex *v2, Vertex *v3 )
{
// Take v1.pos, v2.pos, v3.pos and create a geometric repn..
Triangle tri( v1->pos, v2->pos, v3->pos ) ; // all vertices will
// have to have a .pos member.
// Create the vertex buffer..
VertexBuffer<T> vb ...
}
Then just pass pointers to your inheriting objects (casting as the parent class as necessary)
来源:https://stackoverflow.com/questions/7210944/can-t-in-template-typename-t-use-inheritance