C++ Template: 'is not derived from type'

Question

Why is this code not valid?

#include <vector>

template <typename T>
class A {
  public:
    A() { v.clear(); }

    std::vector<A<T> *>::const_iterator begin() {
      return v.begin();
    }

  private:
    std::vector<A<T> *> v;
};

GCC reports the following errors:

test.cpp:8: error: type 'std::vector<A<T>*, std::allocator<A<T>*> >' is not derived from type 'A<T>'
test.cpp:8: error: expected ';' before 'begin'
test.cpp:12: error: expected `;' before 'private'

What is wrong? How to fix it?

CB Bailey · Accepted Answer · 2014-05-27T22:34:25.277

In this line, you are missing the typename keyword:

std::vector<A<T> *>::const_iterator begin(){

You need:

typename std::vector<A<T> *>::const_iterator begin(){

This because std::vector<A<T> *> is dependent on the template parameter (T) of the class template (A). To enable correct parsing of the template without having to make any assumptions about possible specializations of any other templates, the language rules require you to indicate which dependent names denote types by using the typename keyword.

score 3 · Answer 2 · answered May 15 '10 at 21:10

You need to add typename as the types depend on each other and the compiler can't figure out if it really is a type.

However, on gcc 4.5.0 i get a more concise error message:

test.cc:8:3: error: need ‘typename’ before ‘std::vector<A<T>*>::const_iterator’ because ‘std::vector<A<T>*>’ is a dependent scope

C++ Template: 'is not derived from type'

2 Answers2

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