Question

Here, I test to check a var defined by “typename D”, is a unqualified
dependant name. The answer is YES. since if i write its definition twice:

template<typename D>
class A { 
   D d_field;
   D d_field;

   void func() { d_field = 1000; } 
 }

 int main()
 {
    return 0;
 }

the g++ will report re-definition error. so that it demonstrates “d_field” is checked in template definition.

and if I give a type “double*” to use this template. The template instantiation happens, with an error reported in func2, “int can’t be converted to double*”.

template<typename D>
class A { 
   D d_field;

   void func() { d_field = 1000; } 
 }

 int main()
 {
    A<double*> a; 
    a.func(); 
    return 0;
 }

it prove that “d_field” will be check in template instantiation.

In addition, I change that as below, the g++ will compile it ok.It prove that gcc works in two-phase-lookup based on c++ standard… if the name is qualified dependant name, it will only be checked in template instantiation.

template<typename D>
class A { 
   D:: d_field;
   D:: d_field;

   void func() { d_field = 1000; } 
 }

 int main()
 {
    return 0;
 }

In a word, use a template parameter to define a variable in template, the variable’s name is a unqualified dependant name.

is it right?

Answer 1

No. The variable’s name is a locally declared name, so it is neither dependent nor non-dependent. Only names which the compiler has to look up outside of the template can be dependent or non-dependent.