template class template method specialization

Question

Why does the following code fail to compile?

template <typename T>
struct X
{
    template <typename R>
    R default_value();
};

template <typename T>
int X<T>::default_value<int>()
{
    return -1;
}

it says

x.cpp:17:30: error: template-id ‘default_value<int>’ in declaration of primary template
x.cpp:17:5: error: prototype for ‘int X<T>::default_value()’ does not match any in class ‘X<T>’
x.cpp:13:7: error: candidate is: template<class T> template<class R> R X::default_value()

I also tried to do

template <typename T>
template <>
int X<T>::default_value<int>()
{
    return -1;
}

but this gives me another compilation error

prog.cpp:11: error: invalid explicit specialization before '>' token
prog.cpp:11: error: enclosing class templates are not explicitly specialized
prog.cpp:12: error: template-id 'default_value<int>' for 'int X<T>::default_value()' does not match any template declaration

I also tried doing the same for structures

template <typename T>
struct X
{
    template <typename R> struct default_value;
};

template <typename T>
template <>
struct X<T>::default_value<int>
{
    static int get() { return -1; }
};

same issue.

How to solve that?

why template on the return type, function cannot be overloaded based on return type. — yngccc, Nov 19 '13 at 16:08
I'm not overloading it, I'm doing specialization. Well, but basically I started from structure specialization and encountered the same error. — axe, Nov 19 '13 at 16:11
The same for non-method function works fine. http://ideone.com/C1YTBa — axe, Nov 19 '13 at 16:15

score 2 · Accepted Answer · edited May 23 '17 at 12:19

One cannot explicitly specialize member templates. Consider:

template <class T>
struct X
{
  template <class U> struct Y;
};

...Now (imagine we could do this):

template <class T>
  template <>
  struct X<T>::Y<int>
{};

...For X of which T are we explicitly specializing?

What if, after the point of definition of our explicit specialization, someone does this in one compilation unit...

void foo()
{
  X<int>::Y<int> xy;
}

... and then this in another...(valid code, btw).

template <>
  template<>
  struct X<int>::Y<int>
{};

void foo()
{
  X<int>::Y<int> xy;
}

... which would imply multiple definitions of the same class???

As mentioned previously, this is treated well here

Now, considering that the default value actually depends on the type T, perhaps one can get it from the type T.

template <class T>
struct X
{
  static T defaultValue(){ return T::defaultValue(); }
};

or better yet, one could change the behavior of defaultValue based on whether T has the member defaultValue.

score 0 · Answer 2 · edited May 23 '17 at 12:19

0

template <typename T>
template <>
int X<T>::default_value<int>()
{
    return -1;
}

should be fine. Similar topic...

edited May 23 '17 at 12:19

Community

1
1

answered Nov 19 '13 at 16:01

Naszta

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I tried that already, it says `x.cpp:17:11: error: invalid explicit specialization before ‘>’ token x.cpp:17:11: error: enclosing class templates are not explicitly specialized x.cpp:18:5: error: template-id ‘default_value’ for ‘int X::default_value()’ does not match any template declaration ` – axe Nov 19 '13 at 16:03
@axe: What kind of compiler are you using? – Naszta Nov 19 '13 at 16:21
Locally gcc 4.6.3, The same issue reproduces on ideone.com with gcc 4.8.1 – axe Nov 19 '13 at 16:26

template class template method specialization

2 Answers2