模板中的T可以< typename T>使用继承？

问题描述：

我想做这样的事情：

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 ...
}

由于这不起作用，这是我的解决方法。。

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 ...
}

背景

如果您感兴趣，我正在尝试编写一些代码来处理三角形创建。
每个顶点都必须有一个 .pos 成员，因为每个顶点都必须有一个空格位置。

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.

然而，并非每个顶点类型都具有纹理坐标。并非每个顶点都有颜色。因此参数化类型。

However not every vertex type will have a texture coordinate. Not every vertex will have a color. Hence the parameterized types.

在XNA中使用了类似的方法 VertexBuffer.SetData< T> 。

A similar approach is used in XNA VertexBuffer.SetData<T>.

答

您无法在模板类型参数中指定类型限制。但是，通常情况下，您不需要。

You cannot specify a type restriction in the template type argument. However, generally, you don't have to.

如果您只是这样做：

template <typename T> addTri( T v1, T v2, T v3 )
{
    Vertex &v1r = v1;
    // ....
}

这将有效用Vertex的导数实例化。如果 T& 不可转换为 Vertex& $ c $，它会产生（模糊）错误c>。

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 &.

如果您甚至不关心这些类型是否可转换为 Vertex ，只要它们有相同的成员，你甚至可以跳过作业 - C ++模板参数基本上使用鸭子打字;如果您执行 v1.x ，并且 T 包含名为 x $ c的成员$ c>，然后它会工作，无论 T 可能实际上是什么类型。

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.

你可以多一点使用boost的类型特征库进行复杂化和静态断言;有了这个，你可以开始定义一个断言，使错误更容易理解：

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;
    // ....
}

模板中的T可以< typename T>使用继承？

背景

Background

相关推荐