CODEWITHSUNDEEP
c++c++11lambdaunique-ptrauto
user5110759
user5110759

Reputation:

error: call to implicitly-deleted copy constructor of unique_ptr with auto

I have two template declarations as follows:

template < typename T >
      class node {
        public:
          ...
        const unique_ptr < node < T >> & getParent();
        ...
        private:
          ...
        unique_ptr < node < T >> & P; //for parent node
        ...
      };

    template < typename T1, typename T2 >
      class tree {
        public:
          tree(int, T2 & );
        ...
        void travPreord();
        ...
        private:
          ...
      };

Within the definition of the public method 

template<typename T1, typename T2> tree<T1,T2>::travPreord()

I have the following code:

template < typename T1, typename T2 > 
    void tree < T1, T2 > ::travPreord() {

      //lambda definition

      function < void(unique_ptr < node < T1 >> & ) > prntNode = [ & ]
      (unique_ptr < node < T1 >> & pN) {

        ...

        if(auto rPN = pN - > getParent()) {

          ...

        }

      };

    }

For the assignment inside the if statement condition above, I get the following error from the compiler (g++ 4.2.1):

error: call to implicitly-deleted copy constructor of 'std::__1::unique_ptr >, std::__1::default_delete > > >' if(auto rPN = pN->getParent()){

The arguments to the template instantiation highlighted in the error are supplied from the main():

int main() {

      vector < string > v;

      ...

      tree < string, vector < string >> T(v.size(), v);

      T.travPreord();

      ...

      return 0;

    }

I wrote the if statement condition under question on the following assumptions:

  1. It is possible to assign a unique_ptr to a reference.

  2. The auto keyword should deduce the type of the lvalue expression rPN to be the type of the rvalue expression pN->getParent(), which returns a type unique_ptr<node<T>>&.

So I am not sure about the source of this error.

Could anyone point to the same?

Upvotes: 1

Views: 4966

Answers (1)

songyuanyao
songyuanyao

Reputation: 172924

  1. The auto keyword should deduce the type of the lvalue expression rPN to be the type of the rvalue expression pN->getParent(), which returns a type unique_ptr<node<T>>&.

No, the type will be unique_ptr<node<T>>.

Auto type deduction applies same rules as template argument deduction; the reference part of the expression (i.e. pN->getParent()) is ignored, after that the const part is ignored too. So the result type is unique_ptr<node<T>>.

You need to specify that it's should be a reference explicitly, e.g.

auto& rPN = pN - > getParent() // rPN is of type const unique_ptr<node<T>>&
  1. It is possible to assign a unique_ptr to a reference.

Of course yes. The problem is that rPN is not a reference.

Upvotes: 3

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