为什么我不能将此比较函数作为模板参数传递?

问题描述：

我正在尝试使用我为排序定义的函数创建一个 std::set，但我收到错误:错误:函数GFX::MeshCompare"不是类型名称"

I am trying to create an std::set with a function I defined for sorting, but I get the error: "Error: function "GFX::MeshCompare" is not a type name"

网格.h

namespace GFX
{
    struct Mesh
    {
        [...]
    };

    inline bool MeshCompare(const Mesh& a, const Mesh& b)
    {   
        return ( (a.pTech < b.pTech) ||
                 ( (b.pTech == a.pTech) && (a.pMaterial < b.pMaterial) ) ||
                 ( (b.pTech == a.pTech) && (a.pMaterial == b.pMaterial) && (a.topology < b.topology) ) 
               );
    }
};

渲染器.h

namespace GFX
{
    class Renderer
    {
    private:
        [...]
        std::set<Mesh, MeshCompare> m_Meshes;

    };
};

我做错了什么，我该如何解决?

What am I doing wrong and how do I fix it?

答

std::set 的第二个模板参数必须是 type，而不是价值.

The second template argument to std::set has to be a type, not value .

如果你想使用函数(它是值，而不是类型)，那么你必须将它作为参数传递给构造函数，这意味着你可以这样做这个:

If you want to use function (which is value, not type), then you've to pass it as argument to the constructor, which means you can do this:

class Renderer
{
    typedef bool (*ComparerType)(Mesh const&,Mesh const&);

    std::set<Mesh, ComparerType> m_Meshes;
public:
     Renderer() : m_Meshes(MeshCompare) 
     {        //^^^^^^^^^^^^^^^^^^^^^^^ note this
     }
};

或者，定义一个函子类，并将其作为第二个 type 参数传递给 std::set.

struct MeshComparer
{   
    bool operator()(const Mesh& a, const Mesh& b) const
    {
             return ( (a.pTech < b.pTech) ||
             ( (b.pTech == a.pTech) && (a.pMaterial < b.pMaterial) ) ||
             ( (b.pTech == a.pTech) && (a.pMaterial == b.pMaterial) && (a.topology < b.topology) ) );
   }
};

然后使用它:

std::set<Mesh, MeshComparer> m_Meshes;

为什么我不能将此比较函数作为模板参数传递?

相关推荐