Base class template instantiation depending on derived class constructor argument type

Question

In the following, shouldn't base class constructor be generated by the compiler based on derived class constructor argument type?

template <class T> class foo { int a; public: foo(T a){} // When I convert the constructor to a function template, it works fine. // template <typename T> foo(T a){} }; class bar : public foo<class T> { public: bar(int a):foo(a){} }; int main(void) { bar obj(10); system("pause"); return 0; }

error C2664: 'foo::foo(T)' : cannot convert parameter 1 from 'int' to 'T'

I understand the error, but why is that ?

icecrime · Accepted Answer · 2011-01-06 20:47:01Z

The syntax in class bar : public foo<class T> is incorrect.

Either bar depends on a template parameter T and bar should be a template :

template<class T> class bar : public foo<T> { public: bar(int a):foo(a){} }; int main() { bar<int> obj(10); }

Or you want bar to inherit from a specific instantiation of foo such as :

class bar : public foo<int> { public: bar(int a):foo(a){} }; int main() { bar obj(10); }

From the first code, T in foo<T> say that bar template parameter is also a foo template parameter. Am I correct ?

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