template<typename VertexType>
class CQuad
{
public:
    VertexType tl, bl, tr, br;

    void SetColor(const CColor& color)
    {
        //if VertexType has a member "color", then
        tl.color = color;
        bl.color = color;
        tr.color = color;
        br.color = color;
    }

};

I wrote a template class called Quad with 4 vertex.
If the vertex type has a variable “color”, then I hope to instantiate a SetColor function to fill up all vertices’ color.
If no color variable member for VertexType, just don’t instantiate SetColor function.

I believe it’s possible to do that with SFINAE, but I failed after many trys.
Thanks in advance for your help!

I hope to get the correct code of SFINAE.

/////////////////I tried it out/////////////////////////////

template<typename VertexType, typename CColorType = void>
class CQuad
{
public:
    VertexType tl, bl, tr, br;
};

template<typename VertexType>
class CQuad<VertexType, typename std::enable_if_t<std::is_same<CColor, decltype(VertexType::color)>::value>>
{
public:
    VertexType tl, bl, tr, br;

    void SetColor(const CColor& color)
    {
        tl.color = color;
        bl.color = color;
        tr.color = color;
        br.color = color;
    }
};

But I still feel confused of what I have done.
why cannot
class CQuad<VertexType, typename std::enable_if_t<decltype(VertexType::color)>::value>>?

why enable_if is so important? why can’t remove the keyword typename?

It seems my code is the only answer to run, but I have a strong feeling that we can simplify the code.

Could someone help explaining what is going on?