重载operator - >（）来改变被调用函数的返回值

Question

Basically I'm doing this:

auto result = from (startNodePtr)
    .to<NodeT1>()
    .to<NodeT2>()
    .to<NodeT3>()
    .fail_with(msg)

chasing pointers. You can imagine "from" and "to" returning different templates of a SyntaxSugar class and fail_with returning the result. Works fine.

But it's not that simple. Depending on the types of node I'm on I need to call some operation on the underlying type. like this:

auto result = from(startNodePtr)
    .to<NodeT1>()
    .to<NodeT2>() -> SomeOperation()
    .to<NodeT3>()
    .fail_with(msg)

The operations will in turn return a Pointer, which I again need to wrap in SyntaxSugar.

how would I overload operator-> correctly to do that?

Say my class looks like this:

template<typename Pointer_T>
class SyntaxSugar
{
  Pointer_T ptr;

public:
  explicit SyntaxSugar(Pointer_T ptr_)
    : ptr(ptr_)
  {}

  // all kinds of Syntax Sugar

  // this one I don't get to work:
  auto operator-> () { return SyntaxSugar(ptr.operator->()); }
}

Answer 1

That's amazing, @max-langhof

i basically now do

template<typename Op>
    auto SyntaxSugar::apply(Op func) -> decltype(SyntaxSugar(func(ptr)))
    { 
        if (!ptr) {
            fail();
            return SyntaxSugar(nullptr);
        }
        return SyntaxSugar(func(ptr));
    }

and now I can

auto result = from(startNodePtr)
    .to<NodeT1>()
    .to<NodeT2>()
    .apply([](auto node) { return node->SomeOperation(); })
    .to<NodeT3>()
    .fail_with(msg)