forward declaration of templatized type

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Hello all,
just as everybody would expect following snippet does not compile

C++

namespace NonTemplate{

  struct Unknown;

  struct AnyStruct{
    Unknown *p;
    AnyStruct(Unknown* p): p(p){}
    void doStuff(){ p->doStuff(); } // <-- error: use of undefined type 'NonTemplate::Unknown'
  };

  struct Unknown{
    void doStuff(){}
  };

}

but why this does?

C++

namespace Template {

  template<typename T> struct Unknown;

  template<typename T> struct AnyStruct {
    Unknown<T> *p;
    AnyStruct(Unknown<T>* p) : p(p) {}
    void doStuff() { p->doStuff(); } // <-- OK
  };

  template<typename T> struct Unknown {
    void doStuff() {}
  };
}

P.S. I tried both with VC6 and VC2013.

Posted 11-Jun-15 22:43pm

ucarcamagnu

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Stefan_Lang · Accepted Answer · 2015-06-11T23:37:00

The difference is caused by the fact that for templates the compiler generates the code the moment it is needed!

The compiler doesn't read the entire code before trying to interpret a statement: at this point it will only know the types and definitions it has seen before. Therefore, in your first example, it will not know that the staruct Unknown has a function called doStuff()

In the second case however, the compiler sees a templated class and tries to instantiate that template. As it is scoped within the namespace Template, it will scan the entire namescape for the definition of this template.

The reason the compiler works differently on templates is because it must! Templated code is instantiated only when the compiler finds an actual instantiation, and therefore there is no safe way to ensure all relevant template definitions are parsed in a specific order. Therefore the compiler must be able to search the entire code for template definitions, even in code it hasn't parsed for compilation yet.