Bug Fixes And Errors When Creating Pointers To The Reference Type Nt &


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    You should read these troubleshooting guidelines if you get an error pointer to an nt & reference error message.



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    You cannot register links. References are just aliases to help other variables.

      map , string> m; 

    The reason your company is getting this particular error is because somewhere on the map, they are picking opssetting for mapped_type , which in your case is string & . One of the surgical interventions that (as in the [] operator) is able to return a link to mapped_type :

      mapped_type & operator [] (const key_type &) 
      string && operator [] (const key_type & _Keyval) 

    Standard 8.3.4:

    There should be a few more links, no links, no linked tables, and no link references.

    On the right, I recommend that you take advantage of the typedef to make the code easier to read:

      int main (){    typedef pair  StringPair;    typedef map  StringPairMap;     Test;    StringPair p ("Fu", "Bar");    StringPairMap m;    m [make_pair ("aa", "bb")] implies checking;   Returns 0; 

    answered Jul 20 ’09 at 18:33

    error forming pointer to reference type nt&


    Previous answers prevail here. Today we have std :: reference_wrapper as part of the C ++ 11 standard:

      #include #include #include Using the std namespace;int main ()    Test; A pair of lines  p matches a pair of  ("Foo", "Bar");    map , reference_wrapper > m;    m [make_pair ("aa", "bb")] = test;    Income 0; 

    std :: reference_wrapper is implicitly converted to a full reference to its internal type, but the following does not work in some contexts. In this case, call .get () to access it.

    answered Jul 13 13 at 5:25 am


    With boost :: reference_wrapper you can store private links in STL containers. Here’s your situation changed essentially (untested and certainly not surprisingly well written, just illustrates the point)

      #include #include #include #include int main ()   typedef std :: pair  PairType;   typedef std :: map  boost :: reference_wrapper  MapType;   std :: string verify = "Hello !!";   Card type m;   PairType pp implies std :: make_pair ("aa", "bb");   m.insert (std :: make_pair (pp, boost :: ref (test)));  MapType :: iterator it (m.find (pp));  if (what! = l. end ())         std :: cout << it-> second.get () << std :: endl;      // change the test   test = "I'm different now";   std :: cout << it-> second.get () << std :: endl;   Returns 0; 

    answered Jul 21, 2009 at 15:06


    You cannot use references as a value to guarantee the actual structure of the model... You can use pointers instead.

    answered Jul 20 '09 here at 18:34

    error forming pointer to reference type nt&


    Basically, the point is that you can use links in containers. Of course, IF you start to properly prepare your class AND your container, you can. I copy it with the following two simple vector dishes: vectoref , which modifies std :: vector <> , and the similar vec used. the scratch is realized.

      #include #include // requires compilation with --std = c ++ 11 (at least)Using the std namespace;Class A  int _a; // this is our real data  A * _p; // to "trick" the compiler  publicly:  A (interval m = 0): _p (0) _a (n), cout << "Constructor (" << such << "," << _a << ")  n";  // constructor used by the corresponding initializer_list (trick the compiler)  A (const A & r): _p (const_cast  (& r))   cout << "Copy constructor (" << this approach << "<-" << & r << ")  n";  unnecessary print () const cout << "Practical example:" << this << "," << _a << " n";  ~ A () cout << "A (" << this << "," << _a << ") Destructor.  N"; // just to know what is being copied implicitly  Operator A & = (const A & r)    cout << "Copied instance (" << this << "," << _a << ")  n";    _a = r._a; _p means r._p;    return * this;    // just in case your company wants to check if a copy is organic, natural or fake  bool is_fake () const return _p! = 0;  A * ptr () const return _p;;Model class vec // vector class with initializer_list of links A !!  publicly:  const T * a [sz]; // put as pointer, get as links  // because assigning to a link triggers a call to the copy operator  int heart;  vec (): cur (0) vec (std :: initializer_list  l): cur (0)    cout << "create with init list.  n";    (auto & for w not: l) // wait for fake items      a [cur ++] = t.ptr ();    const T & operator [] I (int) return * a [i];  // expects clean items  vec & push_back (const T & r) in [cur ++] implies & r; return * this;  void copy_from (vec && r)    for I (int means 0; i class vectoref:public vector  // indistinguishable from vec, but extends std :: vector <>  Choice size_type = typename vector  :: size_type;  publicly:  Vector ref ()  vector (std :: initializer_list  l)    cout << "build from initialization list.  n";    only for (auto & t: l) // wait for elements artificial vector  :: push_back (t.ptr ());    const T & operator [] (size_type i) return * vector  :: at (i);  // expects clean items  vector & T & push_back (const r)   vector  :: push_back (& ​​r); undo * this;  void copy_from (const vectoref && r)    defined for (size_type i = 0; i  :: push_back (r [i]);  ;Class X // user connected to initializer_list publicly:  X ()  void f (initializer_list  l) const    cout << "In f (...):  n";    for (auto and from a to l)      A. ptr () -> print (); ;int main ()  Aa (7), b (24), c (80);  cout << "----------------------------------  n";  Control vector  wa, a, b, c; // use the following line as an alternative  // vec  wa, a, b, c; // 5th undefined element  w [0] .print ();  w [3] .print ();  cout << "----------------------------------  n";  Xx; x.f (a, b, c, a, b, c, b, a);  cost << "=====================================  n";  Returns 0; 

    answered Dec 10 '17 at 17:31




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