template and forward declaration

[jobseeker]

From: (e-mail address removed) (jobseeker)
Newsgroups: comp.lang.c++.moderated
Subject: template and forward declaration
NNTP-Posting-Host: 131.233.150.22

I have defined a class that contains a data member of a function type.
The signature of the function type take a pointer of the class itself
as a parameter:

class Softkey; // forward declaration

typedef int (* pfncOnSelect)(Softkey *key, void *);

class Softkey
{

...
int value;
pfncOnSelect m_OnSelect;
...
};

This class is compiled and run successfully without error. The
problem I have now is that I am trying to modify this class into a
templated class as follows:

template<class Type>
class SoftKey; // forward declaration.

typedef int (* pfncOnSelect)(SoftKey *key, void *);

template<class Type>
class Softkey
{

...
Type value;
pfncOnSelect m_OnSelect;
...
};

Now the compiler complains about syntax error in the typedef line.
I have tried with

template<class Type>
typedef int (* pfncDrawCap)(TestKey *key, void *);

or

template<class Type>
typedef int (* pfncDrawCap)(TestKey<Type> *key, void *);

but I still get an error. Can anyone throw some light into this?

 

[=?iso-8859-1?Q?Juli=E1n?= Albo]

jobseeker escribió:

I have tried with

template<class Type>
typedef int (* pfncDrawCap)(TestKey *key, void *);

or

template<class Type>
typedef int (* pfncDrawCap)(TestKey<Type> *key, void *);

but I still get an error. Can anyone throw some light into this?

The language does not allow templatized typedef. You can do something
like:

template <class Type>
struct DrawCap {
typedef int (* pfnc) (TestKey <Type> * key, void *);
};

And use the type DrawCap:fnc.

Regards.

 

[DarkSpy]

jobseeker escribi :




The language does not allow templatized typedef. You can do something
like:

template <class Type>
struct DrawCap {
typedef int (* pfnc) (TestKey <Type> * key, void *);
};

And use the type DrawCap:fnc.

the current is: DrawCap<Type>:fnc; for ex: