Function argument has no type

Question

I came across the following code in the OpenVDB documentation:

template<typename _RootNodeType>
class Tree: public TreeBase
{
...
    template<typename OtherTreeType>
    Tree(const OtherTreeType& other,
        const ValueType& inactiveValue,
        const ValueType& activeValue,
        TopologyCopy): // <-- this looks weird
        TreeBase(other),
        mRoot(other.root(), inactiveValue, activeValue, TopologyCopy())
    {
}

I've seen previously that an argument defaults to an int if no type is specified, but could this be the case here? TopologyCopy is being called as an operator 2 lines below.

What does the above declaration do/mean?

Edit: The accepted answer explains what is happening. The solution is to call the function as

openvdb::Tree newTree(oldTree, inactiveValue, activeValue, TopologyCopy());

Looks weird because it IS weird. Why would someone take an argument, not use it, and default construct the same type?? I'd hope somewhere there's a comment as to why or an obvious reason that argument needs to be there to distinguish it from some other version of the constructor. — Edward Strange, Jul 22 '16 at 00:00

score 3 · Accepted Answer · answered Jul 21 '16 at 23:58

3

It's not an argument without a type. It's an argument without a name. Its type is TopologyCopy. And TopologyCopy() is default constructing an object of that type and passing it to the constructor of mRoot. If I had to guess, I would say they are probably using tag dispatching here to select between different constructors with otherwise identical arguments.

answered Jul 21 '16 at 23:58

Benjamin Lindley

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Hmm, that makes sense. Do you know how would I go about calling that constructor? Using only 3 arguments results in a compiler error. – pingul Jul 22 '16 at 00:03
3

@pingul: Pass `TopologyCopy()` as the fourth argument, just like they did for `mRoot`. – Benjamin Lindley Jul 22 '16 at 00:04

Jarod42 · Answer 2 · 2016-07-22T00:04:17.897

0

TopologyCopy is a type, and as the argument/variable is not used, it is not present.

Following TopologyCopy() constructs a TopologyCopy.

edited Jul 22 '16 at 00:04

answered Jul 21 '16 at 23:57

Jarod42

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You can't really *call* a constructor as a user, since constructors don't have a name. And imagine `using TopologyCopy = int;` – Kerrek SB Jul 22 '16 at 00:00

Function argument has no type

2 Answers2