base class public type (non template and template base class)

[Hicham Mouline]

Hello,

This compiles

#include <iostream>

class Base {
public:
typedef double type;
};

class Derived : public Base {
public:
void f(type t) const {}
};


While this

template <typename T>
class Base {
public:
typedef double type;
};

template <typename T>
class Derived : public Base<T> {
public:
void f(type t) const {} // type has not been declared
};


This templated version requires

template <typename T>
class Derived : public Base<T> {
public:
void f(typename Base<T>::type t) const {}
};


Is g++3.4 standard compliant here? Or is this a g++3.4 restriction?

writing
typename Base<T>::type
is tedious?

I tried to put type in a non template base class B of Base, but in Derived,
I still needed to qualify B::type

rds,

 

[Victor Bazarov]

This compiles

#include <iostream>

class Base {
public:
typedef double type;
};

class Derived : public Base {
public:
void f(type t) const {}
};


While this

template <typename T>
class Base {
public:
typedef double type;
};

template <typename T>
class Derived : public Base<T> {
public:
void f(type t) const {} // type has not been declared
};


This templated version requires

template <typename T>
class Derived : public Base<T> {
public:
void f(typename Base<T>::type t) const {}
};


Is g++3.4 standard compliant here? Or is this a g++3.4 restriction?

writing
typename Base<T>::type
is tedious?

I tried to put type in a non template base class B of Base, but in Derived,
I still needed to qualify B::type

FAQ section 35. See "dependent name".

V