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Home/ Questions/Q 793635
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T22:11:01+00:00 2026-05-14T22:11:01+00:00

Why is this code not valid? #include <vector> template <typename T> class A {

  • 0

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?

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  1. Editorial Team
    Editorial Team
    2026-05-14T22:11:02+00:00Added an answer on May 14, 2026 at 10:11 pm

    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.

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