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jobseeker
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 Posted: Tue Oct 14, 2003 9:21 pm Post subject: template and forward declaration |
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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?
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Ben Hutchings
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| Posted: Thu Oct 16, 2003 2:10 pm Post subject: Re: template and forward declaration |
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In article <6bb7c82b.0310141309.40ba4c47 (AT) posting (DOT) google.com>,
jobseeker wrote:
<snip>
| Quote: | 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 Softkey
{
...
Type value;
pfncOnSelect m_OnSelect;
...
};
Now the compiler complains about syntax error in the typedef line.
|
Right, SoftKey is not a type name.
| Quote: | I have tried with
template
typedef int (* pfncDrawCap)(TestKey *key, void *);
or
template
typedef int (* pfncDrawCap)(TestKey
but I still get an error. Can anyone throw some light into this?
|
There is no such thing as a template typedef. Also, you have
inconsistently changed names in your code.
In the absence of a template typedef or similar feature, you can
use either this:
template<class Type>
class SoftKey
{
typedef int (* pfncOnSelect)(SoftKey *key, void *);
Type value;
pfncOnSelect m_OnSelect;
};
(replacing pfncOnSelect with SoftKey<Type>::pfncOnSelect)
or this:
template<class Type>
class SoftKey;
template<class Type>
struct pfncOnSelect
{
typedef int (* type)(SoftKey<Type> *key, void *);
};
template<class Type>
class SoftKey
{
Type value;
typename pfncOnSelect<Type>::type m_OnSelect;
};
(replacing pfncOnSelect with pfncOnSelect<Type>::type).
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Siemel Naran
Guest
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| Posted: Thu Oct 16, 2003 2:22 pm Post subject: Re: template and forward declaration |
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"jobseeker" <jobseeker95479 (AT) yahoo (DOT) com> wrote in message
| Quote: | template<class Type
class SoftKey; // forward declaration.
typedef int (* pfncOnSelect)(SoftKey *key, void *);
template
typedef int (* pfncDrawCap)(TestKey *key, void *);
|
In the above and below, what is 'TestKey'? Did you mean 'SoftKey'?
| Quote: | | template
typedef int (* pfncDrawCap)(TestKey
|
The language does not support template typedefs. There is a close
workaround though: a template struct, with a typedef in it.
template<class Type>
struct pfncDrawCap {
typedef int (* FunctionPtr)(SoftKey<Type> *key, void *);
};
And then
template<class Type>
class Softkey
{
...
Type value;
typename pfncOnSelect<Type>::FunctionPtr m_OnSelect;
...
};
You could perhaps make struct pfncDrawCap a nested struct, that is
Softkey::pfncDrawCap.
--
+++++++++++
Siemel Naran
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Ron
Guest
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| Posted: Thu Oct 16, 2003 2:28 pm Post subject: Re: template and forward declaration |
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| Quote: | 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 SoftKey; // forward declaration.
typedef int (* pfncOnSelect)(SoftKey *key, void *);
template
class Softkey
{
...
Type value;
pfncOnSelect m_OnSelect;
...
};
Now the compiler complains about syntax error in the typedef line....
|
Yep. C++ doesn't support templated typedefs. One alternative would be
to derive SoftKey from a class that supports all the methods you need,
then override those methods in SoftKey. Then you could write you
typedef as
typedef int (* pfncOnSelect)(SoftKeyBase *keyBase, void *);
-Ron
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Christoph Schulz
Guest
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| Posted: Thu Oct 16, 2003 2:31 pm Post subject: Re: template and forward declaration |
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Hello!
jobseeker <jobseeker95479 (AT) yahoo (DOT) com> wrote:
| Quote: | 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
{
[snip]
};
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 *);
|
This doesn't work because Softkey does not denote a class but a
class template.
| Quote: | [snip]
Now the compiler complains about syntax error in the typedef line.
I have tried with
template
typedef int (* pfncDrawCap)(TestKey *key, void *);
or
template
typedef int (* pfncDrawCap)(TestKey
but I still get an error. Can anyone throw some light into this?
|
You can't name a type via typedef which depends on a template parameter.
So any use of "template <...> typedef ..." is illegal.
You can fix your example by moving the typedef into the template class,
where it is apparently used to define the member m_onSelect:
template<class Type>
class Softkey
{
typedef int (* pfncOnSelect)(Softkey *key, void *);
Type value;
pfncOnSelect m_OnSelect;
}
This works because "Softkey" in the typedef is really "Softkey<Type>".
So you defined a distinctive typedef for each possible Softkey
specialization. Softkey<int>::pfncOnSelect is different from
Softkey<char>::pfncOnSelect, for instance. But that's understandable,
because Softkey<int> is not Softkey<char> either.
To summarize, every time you need a templated typedef you can move
it into a templated class. E.g. if you need something like:
template <class T1, class T2> typedef T1 (*myTypedef) (T2);
you'll need to write:
template <class T1, class T2> struct myTypedef {
typedef T1 (*type) (T2);
};
and you can use it as follows:
myTypedef <int, char>::type callback;
Regards,
Christoph
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Tony Oliver
Guest
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| Posted: Thu Oct 16, 2003 5:28 pm Post subject: Re: template and forward declaration |
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[email]jobseeker95479 (AT) yahoo (DOT) com[/email] (jobseeker) wrote
| Quote: | 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 Softkey
{
...
Type value;
pfncOnSelect m_OnSelect;
...
};
Now the compiler complains about syntax error in the typedef line.
I have tried with
template
typedef int (* pfncDrawCap)(TestKey *key, void *);
or
template
typedef int (* pfncDrawCap)(TestKey
but I still get an error. Can anyone throw some light into this?
|
You're nearly there. Unfortunately, you (like the rest of us) want
templated typedefs, which are currently not part of the language (but
I believe this is an issue under discussion for possible inclusion in
the next Standard).
There is a well-known trick to circumvent this limitation (it's even
used in the Standard Template Library). You have to wrap your typedef
inside a templated structure as follows:
template<class Type>
class SoftKey; // forward declaration.
template<class Type>
struct Specialised_On
{
typedef int (*pfncOnSelect)(SoftKey<Type> *key, void *);
};
template<class Type>
class Softkey
{
//...
Type value;
Specialised_On<Type>::pfncOnSelect m_OnSelect;
//...
};
Hope this helps.
Best regards,
Tony.
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Sebastian Faust
Guest
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| Posted: Fri Oct 17, 2003 2:29 am Post subject: Re: template and forward declaration |
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Hi,
try the follwing:
template<class Type>
class SoftKey; // forward declaration.
typedef int (* pfncOnSelect)(SoftKey<class Type> *key, void *);
template<class Type>
class Softkey
{
Type value;
pfncOnSelect m_OnSelect;
};
Bye,
Sebastian
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Thomas Mang
Guest
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| Posted: Sun Oct 19, 2003 9:42 pm Post subject: Re: template and forward declaration |
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Tony Oliver schrieb:
| Quote: | You're nearly there. Unfortunately, you (like the rest of us) want
templated typedefs, which are currently not part of the language (but
I believe this is an issue under discussion for possible inclusion in
the next Standard).
There is a well-known trick to circumvent this limitation (it's even
used in the Standard Template Library). You have to wrap your typedef
inside a templated structure as follows:
template<class Type
class SoftKey; // forward declaration.
template
struct Specialised_On
{
typedef int (*pfncOnSelect)(SoftKey
};
template<class Type
class Softkey
{
//...
Type value;
Specialised_On
|
Almost correct. This line should start with a "typename", since it depends on
a template parameter:
typename Specialised_On<Type>::pfncOnSelect m_OnSelect;
regards,
Thomas
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Tony Oliver
Guest
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| Posted: Mon Oct 20, 2003 10:03 pm Post subject: Re: template and forward declaration |
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Thomas Mang <a9804814 (AT) unet (DOT) univie.ac.at> wrote
| Quote: | Tony Oliver schrieb:
SNIP
template<class Type
class Softkey
{
//...
Type value;
Specialised_On
Almost correct. This line should start with a "typename", since it depends on
a template parameter:
typename Specialised_On<Type>::pfncOnSelect m_OnSelect;
//...
};
regards,
Thomas
|
D'oh! Thanks, Thomas.
I forgot to re-test this under g++; I must have (uncharacteristically)
only tested this under MSVC6, which fails to warn about the deprecated
nature of compiler-deduced types taken from template arguments.
Typical...
Tony.
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