Question

template <class T>
class Test{

public:
  Test(T){ }

  template <class U>
  U to() const{ return U(0); }
};


template <class Real>
class Base{

  typedef Test<Real> TST;

protected:
  Base(const TST& t){

    _i = t.to<int>();  // <- but this gives error
  }

  int _i;
};


template <class Real, class T> class Child;

template <class Real>
class Child<Real, int> : public Base<Real>{

  typedef Base<Real> F;
  typedef Test<Real> TST;

public:
  Child() : F(TST(23.0)){ }
};



int main(int argc, char* argv[]){

  Child<double, int> rw;

  Test<double> t1(3.3);
  int i = t1.to<int>();   // <- this works

  return 0;
}

Calling to in the main works, but I can’t figure out why it doesn’t when called in Base(). The error I get is:

t1.cpp: In constructor ‘Base<Real>::Base(const TST&)’:
t1.cpp:20:15: error: expected primary-expression before ‘int’
t1.cpp:20:15: error: expected ‘;’ before ‘int’

Answer 1

Because t is of type Test<Real> where Real is a template argument, so the type of t is actually a dependent type. You need to let the compiler know to is a template function by using:

_i = t.template to<int>();

Otherwise you could be attempting to use the operator< on a member of t called to (causing some errors), which is what the compiler assumes unless you mark it template. When called from main, the type of t1 is Test<double> which is not dependent on any template argument so the compiler can figure on its own its a template function.