CODEWITHSUNDEEP
c++listunique-ptr
Antoine C.
Antoine C.

Reputation: 3952

Typedef of unique_ptr List

I'd like to make a typedef for std::list<std::unique_ptr<>>, so I could type PtrList<A> and it would be replaced by std::list<std::unique_ptr<A>>.

I know the way to do that with #define:

#define PtrList(x) std::list<std::unique_ptr<x>>

But I think typedef would be proper. However I don't know how to realize that. Could someone tell me how to typedef this ?

Upvotes: 2

Views: 2116

Answers (4)

Arne Mertz
Arne Mertz

Reputation: 24576

In C++11: use a template alias

template <class T>
using PtrList = std::list<std::unique_ptr<T>>;

//later:
PtrList<int> myIntPtrList;

in C++98/03: There is no template alias, so in cases like this you will need a workaround. There is no std::unique_ptr either, so you obviously have C++11 already, but for the sake of completeness here is a C++03 example with a shared_ptr (i refuse to use auto_ptr, regardless of it is "more like unique_ptr"):

template <class T>
struct PtrList { typedef std::list<boost::shared_ptr<T>> type; };

//later:
PtrList<int>::type myIntPtrList;

Upvotes: 3

Wojtek Surowka
Wojtek Surowka

Reputation: 20993

In the current version of C++ (C++11) you can using type aliases:

template<class T> using PtrList = std::list<std::unique_ptr<T>>;

Upvotes: 2

utnapistim
utnapistim

Reputation: 27365

First, try not to use #define to generate code. Except for conditional compilation, #define should be avoided.

A correct type alias:

template<class T> using PtrList = std::list<std::unique_ptr<T>>;

Example use:

PtrList<int> intPtrList;

Upvotes: 9

Some programmer dude
Some programmer dude

Reputation: 409136

How about

template<typename A>
using PtrList = std::list<std::unique_ptr<A>>;

PtrList<SomeStructure> myList;

Upvotes: 5

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