CODEWITHSUNDEEP
c++c++17
Jimmy Zoto
Jimmy Zoto

Reputation: 1763

Linkage of alias that requires decltype

I have a class member that is std::unique_ptr with a custom deleter. The custom deleter is a lambda function around a c function from an external library I do not control. I want to alias the unique_ptr type so that clients don't have to be exposed to the full signature. That is, I want :

struct {
   mytype_ptr foo;
}

instead of:

struct {
   std::unique_ptr<mytype, lambda<blah>> foo;
}

I tried doing this in the class header file:

auto x = [](mytype* t){ mytype_destroy(t); };
using mytype_ptr = std::unique_ptr<mytype, decltype(x)>;

but of course the linker complains about multiple definitions.

I have tried various permutations, including inlining the lambda definition in decltype() and others, without success.

Is there a clean way to do this?

Upvotes: 0

Views: 125

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 473946

No, you cannot. lambda<blah> is not an implementation detail of the object; it is part of the object. The deleter of a unique_ptr is part of its type. And this type is a member of the object. So any external users of that type will have to know about all of its members.

Unless you use PIMPL or type erasure (via std::shared_ptr, for example) or something similar, external users will need to access the type.

If all you want to do is keep users from having to know about the C API, you can write a simple function declaration or functor in the header, and put the implementation of its deletion operation in a source file:

//In header
struct my_deleter
{
  void operator()(mytype*) const;
};

Upvotes: 2

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