CODEWITHSUNDEEP
c++templatesshared-ptr
gatherer
gatherer

Reputation: 43

constructing templated containers that are themselves shared_ptr types

I'm having some difficulty wrapping my head around this one, likely because I don't fully understand the underlying mechanics.

I have a struct like the following:

template <class T>
struct A {
    int id;
    T x;
}

This struct is then used as a type for a shared_ptr:

typedef std::shared_ptr<A> A_ptr;

Which in turn gets stored in a map:

typedef unordered_map<int , A_ptr> AMap;

However, when I compile, I get the following error:

> error: type/value mismatch at argument 1 in template parameter list
> for 'template<class _Tp> class std::shared_ptr'   typedef
> std::shared_ptr<A> A_ptr;

I have read other posts about a similar issue, but all of them recommend creating the shared_ptr A inside the struct. I've tried this, even though I can't logically understand why it would help with my problem, but it doesn't work. Any help is appreciated.

Upvotes: 0

Views: 49

Answers (1)

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170074

A is the name of a template, not the name of a class. If you were to name a specialization, your typedef would be fine:

typedef std::shared_ptr<A<int>> A_int_ptr; // A<int> **is** a class

If you want to keep the code generic, you'll need to switch to using alias templates instead of regular type aliases:

template<typename T>
using A_ptr = std::shared_ptr<A<T>>;

template<typename T>
using AMap = std::unordered_map<int , A_ptr<T>>;

Now everything is properly templated, and the instantiations depend on the type T you intend to pass onto A.

Upvotes: 8

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