Recursive alias template

Question

I'm creating an adjacency list to modelize an oriented weighted graph, and i'd like to know if recursive alias templates are possibe. I've tried something like this

template<class edge_data>
using graph_test = std::set<std::map<std::set<graph_test>::const_pointer, edge_data>>;

(each entry of the set is a map, and each map represents a vertex. each entry of the maps are edges: keys are pointers to other vertices, values are weight. the pointers are not invalidated on insertion)

This produces an error : 'graph_test': undeclared identifier.

I know I could use std::any to avoid this problem, but it look a bit 'dangerous', even if I know what I'm doing.

Do you have any suggestions/ideas that might solve this problem ?

EDIT 1: As Igor Tandetnik pointed it out, the way I create my const_pointer could not be correct. I came up with this code (modifications in italic):

template<class edge_data>
    using graph_test = std::set<std::map<graph_test<edge_data>::const_pointer, edge_data>>;

The problem and the error are still here, but it's a least one mistake fixed.

EDIT 2: I took a look at bipll's solution, which was very relevant, and Dani's comment, showing a probable lack of typedef keyword. I came up with this piece of code, and it compiles:

template<class edge_data> struct graph_test
    : std::set<typename std::map<typename graph_test<edge_data>::const_pointer, edge_data>> {
    using std::map<typename graph_test<edge_data>::const_pointer, edge_data>::map;
};

I will mark with question as solved, thank you very much for your time and help :)

`std::set` expects `T` to be a type, but `graph_test` is not a type - it's a class template, an infinite family of types. For that reason alone, `std::set` already doesn't make sense. — Igor Tandetnik, Jul 01 '18 at 13:35
Possible duplicate of [How to define a recursive type?](https://stackoverflow.com/questions/25563661/how-to-define-a-recursive-type) — cpplearner, Jul 01 '18 at 13:57
Well it's the same kind of problem, but I'm using an alias template instead of a typedef, so it wouldn't have helped a lot. But you're a bit of right, it kind of looks similar. — Antoine, Jul 01 '18 at 14:15

bipll · Accepted Answer · 2018-07-01T16:27:20.170

0

In your definition, graph_test lacks a template parameter, and a typename is missing. Besides, a type cannot be an alias to itself, its expansion would create an infinite dependency graph. You can make it a real type, not an alias:

template<class edge_data> struct graph_test
    :std::set<
        typename std::map<typename std::set<graph_test<edge_data>>::const_pointer, edge_data>>
{
    using std::map<typename std::set<graph_test<edge_data>>::const_pointer, edge_data>>::map;
};

edited Jul 01 '18 at 16:27

answered Jul 01 '18 at 13:58

bipll

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Thanks, your answer looks great :) However I got many errors when I tried it, do you think it could come from the compilers arguments ? – Antoine Jul 01 '18 at 14:02
I think you're missing typename before `std::set>::const_pointer` – Daniel Jul 01 '18 at 14:03

Recursive alias template

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