C++ boost::ptr_vector::iterator problem

Question

I have this class:

template <class S, class P, class A>
class Task
{
  private:

    timeval start;
    boost::ptr_vector<S> states;
    boost::ptr_vector<P> policies;

  public:

    P findPolicy(S *state);
    S findState(S *state);

};

When I try to define findPolicy or findState using an iterator:

template <class S, class P, class A>
S Task<S,P,A>::findState(S *state)
{
  boost::ptr_vector<S>::iterator it;
  for ( it = policies.begin(); it < policies.end(); ++it)
  {
    // blah 
  }
}

Defined after the class, compiler says:

error: expected ';' before it;

Even trying to define the function in the class declaration gives me the same error. I'm puzzled since this is how I've been using boost::ptr_vector iterators so far. The only thing that seems to work is good old fashioned:

for (int i = 0; i < policies.size(); i++)
  {
    if (policies[i].getState() == state)
    {
     return policies[i];
    }
  }

What am I missing ?

It's always a good idea, when posting an error, to point out the line that the error appears on. — Nicol Bolas, Jul 14 '11 at 07:14
Don't use "it < policies.end()". Correct is "it != policies.end()" — Konstantin Tenzin, Jul 14 '11 at 09:21

Nicol Bolas · Accepted Answer · 2011-07-14T08:28:50.663

7

  boost::ptr_vector<S>::iterator it;

This needs to use the C++ keyword typename:

typename boost::ptr_vector<S>::iterator it;

Otherwise, C++ doesn't know what ptr_vector<S>::iterator is supposed to be. This is because the definition of ptr_vector<S> depends on the template parameter S, and the value of S is not known at the time of the template's definition. But the compiler needs to be able to make sense of the line ptr_vector<S>::iterator without knowing what S is exactly.

So compilers assume that dependent names are variables (so a static member of ptr_vector<S>); you need to use typename in order to tell the compiler that the dependent name is a type and not a variable.

edited Jul 14 '11 at 08:28

answered Jul 14 '11 at 07:13

Nicol Bolas

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@Alex: The problem is that `boost::ptr_vector` is dependent on template parameter `S` and so the compiler has no way of knowing whether it is a type or a value. When such an ambiguity arises, it assumes it is value, which then leads to compilation errors. Qualifying the declaration with the `typename` keyword tells the compiler to treat it as a dependent type. – Praetorian Jul 14 '11 at 07:36

score 4 · Answer 2 · answered Jul 14 '11 at 07:15

I think you need to add typename - I think you need:

typename boost::ptr_vector<S>::iterator it;

Whenever with templates you see some error like:

error: expected ';' before it;

It is because the type in front of it is not being seen by the compiler as a type and so one needs to add a typename.

score 0 · Answer 3 · answered Jul 14 '11 at 07:14

0

It doesn't explain the error message IMO, but in the code you posted you assign a boost::ptr_vector<S>::iterator variable with a boost::ptr_vector<P>::iterator value.

answered Jul 14 '11 at 07:14

themel

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@Alex - In `findState` you use a `vector~~::iterator` when `policies` is a `vector~~
`. Of course this depends on the actual S and P, so sometimes it might be ok.
– Bo Persson Jul 14 '11 at 07:49

C++ boost::ptr_vector::iterator problem

3 Answers3