Operator overloading on class templates

Question

I m having some problems defining some operator overloads for template classes  Let s take this hypothetical class for example   template  lt class T gt  class MyClass                  operator       In MyClass h MyClass lt T gt  amp  operator   const MyClass lt T gt  amp  classObj        In MyClass cpp template  lt class T gt  MyClass lt T gt  amp  MyClass lt T gt   operator   const MyClass lt T gt  amp  classObj               return  this      Results in this compiler error   no match for  operator    in  classObj2    classObj1   operator lt  lt      In MyClass h friend std  ostream amp  operator lt  lt  std  ostream amp  out  const MyClass lt T gt  amp  classObj        In MyClass cpp template  lt class T gt  std  ostream amp  operator lt  lt  std  ostream amp  out  const MyClass lt T gt  amp  classObj                   return out      Results in this compiler warning   friend declaration  std  ostream amp  operator lt  lt  std  ostream amp   const MyClass lt T gt  amp    declares a non-template function    What am I doing wrong here

User · Answer

This helped me with the exact same problem.

Solution:

Forward declare the friend function before the definition of the class itself. For example:

   template<typename T> class MyClass;  // pre-declare the template class itself
   template<typename T> std::ostream& operator<< (std::ostream& o, const MyClass <T>& x);

Declare your friend function in your class with "<>" appended to the function name.
```
   friend std::ostream& operator<< <> (std::ostream& o, const Foo<T>& x);
```

User · Answer

In MyClass h MyClass lt T gt  amp  operator   const MyClass lt T gt  amp  classObj        In MyClass cpp template  lt class T gt  MyClass lt T gt  amp  MyClass lt T gt   operator   const MyClass lt T gt  amp  classObj               return  this      This is invalid for templates  The full source code of the operator must be in all translation units that it is used in  This typically means that the code is inline in the header   Edit  Technically  according to the Standard  it is possible to export templates  however very few compilers support it  In addition  you CAN also do the above if the template is explicitly instantiated in MyClass cpp for all types that are T- but in reality  that normally defies the point of a template   More edit  I read through your code  and it needs some work  for example overloading operator    In addition  typically  I would make the dimensions part of the template parameters  allowing for the failure of   or    to be caught at compile-time  and allowing the type to be meaningfully stack allocated  Your exception class also needs to derive from std  exception  However  none of those involve compile-time errors  they re just not great code

User · Answer

This way works    class A       struct Wrap               A amp  a          Wrap A amp  aa  aa a             operator int     return a value            operator std  string     stringstream ss  ss  lt  lt  a value  return ss str                 Wrap operator      return Wrap  this

User · Answer

You must specify that the friend is a template function   MyClass lt T gt  amp  operator   lt  gt  const MyClass lt T gt  amp  classObj     See this C   FAQ Lite answer for details

User · Answer

You need to say the following  since you befriend a whole template instead of just a specialization of it  in which case you would just need to add a  lt  gt  after the operator lt  lt     template lt typename T gt  friend std  ostream amp  operator lt  lt  std  ostream amp  out  const MyClass lt T gt  amp  classObj     Actually  there is no need to declare it as a friend unless it accesses private or protected members  Since you just get a warning  it appears your declaration of friendship is not a good idea  If you just want to declare a single specialization of it as a friend  you can do that like shown below  with a forward declaration of the template before your class  so that operator lt  lt  is regognized as a template      before class definition     template  lt class T gt  class MyClass      note that this  T  is unrelated to the T of MyClass   template lt typename T gt  std  ostream amp  operator lt  lt  std  ostream amp  out  const MyClass lt T gt  amp  classObj       in class definition     friend std  ostream amp  operator lt  lt   lt  gt  std  ostream amp  out  const MyClass lt T gt  amp  classObj     Both the above and this way declare specializations of it as friends  but the first declares all specializations as friends  while the second only declares the specialization of operator lt  lt  as a friend whose T is equal to the T of the class granting friendship    And in the other case  your declaration looks OK  but note that you cannot    a MyClass lt T gt  to a MyClass lt U gt  when T and U are different type with that declaration  unless you have an implicit conversion between those types   You can make your    a member template     In MyClass h template lt typename U gt  MyClass lt T gt  amp  operator   const MyClass lt U gt  amp  classObj        In MyClass cpp template  lt class T gt  template lt typename U gt  MyClass lt T gt  amp  MyClass lt T gt   operator   const MyClass lt U gt  amp  classObj               return  this

[c++] Operator overloading on class templates

Examples related to c++

Examples related to templates

Examples related to operator-overloading