是否将一个具有空dtor的类在联合中显式调用它自己的类时称为成员的dtor？

这很难说，希望伪代码更直接。在此示例中，Texture::~Texture()会隐式调用source.bitmap.pixels.~vector()吗？

struct Bitmap{
    ~Bitmap(){} // empty dtor

    // members
    std::vector<uint8> pixels;  // <-- will this dealloc when ~Bitmap() is called manually?
};

struct Texture{
    ~Texture(){
        // assume sourceType is 1
        switch(sourceType){ 
        case 1:
            source.bitmap.~Bitmap();
            break;
        }
    }

    // members
    uint sourceType;
    union Source{
        Source(){}
        ~Source(){}

        // members
        Bitmap bitmap;
    }source;
};