By using this site, you agree to our updated Privacy Policy and our Terms of Use. Manage your Cookies Settings.
443,369 Members | 1,138 Online
Bytes IT Community
+ Ask a Question
Need help? Post your question and get tips & solutions from a community of 443,369 IT Pros & Developers. It's quick & easy.

Passing pointer to template function as argument to pointer to template function

P: n/a
I was decomposing a task into different policies. Essentially, there is
a general option obtained from a server and user options obtained from
configuration variables. The two options are complementary to one
another.

So I proceeded to decompose the tasks that deal with user options into
two functions. Each of the functions do something and towards the end
they do supplementary tasks that depend on the server option.

The whole things fits in a framework which notifies listeners before
and after the main tasks and so the notification scheme for the
listeners are separate policies as well.

The whole thing is combinatorially complex and hence my decision to
split things up using templates as policies. Here is where I run into
my question: It is exemplified by the following simple code:

#include <iostream>

// main task function
template<class FuncPtr>
void Foo(FuncPtr ptr)
{
ptr();
}

// want to pass an instance of this to Foo
template<class X>
void Bar(X x)
{
x();
}

// want to pass pointer to this to Bar
void G()
{
std::wcout << "From G!" << std::endl;
}

int main()
{
Foo(&Bar(&G)); // doesn't work, coz Bar(&G) is a call to
// function Bar with pointer to G and
// inferred template argument

Foo(&Bar<G>); // don't work becuase the compiler could
// not infer the template argument
// type for overloadl function type from
// overloaded function type

Foo< Bar<G> >( &Bar<G> ); // doesn't work I suppose because
// I should pass the template type as
// pointer to Bar<G> but the compiler
// complains as "invalid template
// argument for FuncPtr; type
// expected"

typedef void (*GPtr) ();
typedef void (*BarGPtr)(GPtr) ;
Foo< BarGPtr >( &Bar<G> ); // doesn't work; compiler complains
// that param 1 for Foo cannot be
// converted from type void(T1) to
// BarGPtr. I suppose T1 is a temporary
// type?
return 0;
}

Any ideas on this?

thanks,

-vijai.

Nov 8 '05 #1
Share this Question
Share on Google+
4 Replies


P: n/a
Vijai Kalyan wrote:
I was decomposing a task into different policies. Essentially, there is
a general option obtained from a server and user options obtained from
configuration variables. The two options are complementary to one
another.

So I proceeded to decompose the tasks that deal with user options into
two functions. Each of the functions do something and towards the end
they do supplementary tasks that depend on the server option.

The whole things fits in a framework which notifies listeners before
and after the main tasks and so the notification scheme for the
listeners are separate policies as well.

The whole thing is combinatorially complex and hence my decision to
split things up using templates as policies. Here is where I run into
my question: It is exemplified by the following simple code:

#include <iostream>

// main task function
template<class FuncPtr>
void Foo(FuncPtr ptr)
{
ptr();
}

// want to pass an instance of this to Foo
template<class X>
void Bar(X x)
{
x();
}

// want to pass pointer to this to Bar
void G()
{
std::wcout << "From G!" << std::endl;
}

int main()
{
Foo(&Bar(&G)); // doesn't work, coz Bar(&G) is a call to
// function Bar with pointer to G and
// inferred template argument

Foo(&Bar<G>); // don't work becuase the compiler could
// not infer the template argument
// type for overloadl function type from
// overloaded function type

Foo< Bar<G> >( &Bar<G> ); // doesn't work I suppose because
// I should pass the template type as
// pointer to Bar<G> but the compiler
// complains as "invalid template
// argument for FuncPtr; type
// expected"

typedef void (*GPtr) ();
typedef void (*BarGPtr)(GPtr) ;
Foo< BarGPtr >( &Bar<G> ); // doesn't work; compiler complains
// that param 1 for Foo cannot be
// converted from type void(T1) to
// BarGPtr. I suppose T1 is a temporary
// type?
return 0;
}

Any ideas on this?

thanks,

-vijai.


How about using a functor instead (only changed code is shown):

template<class X>
struct Bar
{
X x_;
Bar( const X& x ) : x_( x ) {}
void operator()()
{
x_();
}
};

int main()
{
typedef void (*GPtr)();
Foo( Bar<GPtr>(G) );
return 0;
}

Cheers! --M

Nov 8 '05 #2

P: n/a
Yep, I thought about a functor, but I am not sure about using it purely
from an aesthetic perspective. But thanks anyway. If nobody comes up
with something then I will use a functor, but I still hope someone will
know what the syntax is ... I googled it, but came up with pageful of
results about function templates but not of pointers to them. It sure
would be nice to know!

regards,

-vijai.

Nov 8 '05 #3

P: n/a
"Vijai Kalyan" <vi**********@gmail.com> wrote in message
news:11**********************@o13g2000cwo.googlegr oups.com
I was decomposing a task into different policies. Essentially, there
is a general option obtained from a server and user options obtained
from configuration variables. The two options are complementary to one
another.

So I proceeded to decompose the tasks that deal with user options into
two functions. Each of the functions do something and towards the end
they do supplementary tasks that depend on the server option.

The whole things fits in a framework which notifies listeners before
and after the main tasks and so the notification scheme for the
listeners are separate policies as well.

The whole thing is combinatorially complex and hence my decision to
split things up using templates as policies. Here is where I run into
my question: It is exemplified by the following simple code:

#include <iostream>

// main task function
template<class FuncPtr>
void Foo(FuncPtr ptr)
{
ptr();
}

// want to pass an instance of this to Foo
template<class X>
void Bar(X x)
{
x();
}


In that case, Foo is defined wrongly. If you pass an instance of Bar to Foo,
then ptr will be &Bar<something> so

ptr();

translates into

Bar<something>();

which is wrong because Bar<something> has an X parameter. You must call:

ptr(x);

However, your Foo function isn't set up to be able to make sense of this x
argument. Try this

template<class BarFuncPtr, class GFunctPtr>
void Foo(BarFuncPtr bptr, GFunctPtr gptr)
{
bptr(gptr);
}

// want to pass an instance of this to Foo
template<class X>
void Bar(X x)
{
x();
}

// want to pass pointer to this to Bar --- via Foo
void G()
{
std::wcout << "From G!" << std::endl;
}

int main()
{
typedef void (*GPtr) ();
Foo(&Bar<GPtr>, &G);
return 0;
}
--
John Carson

Nov 8 '05 #4

P: n/a
That's interesting. So, this means that Foo has to have knowledge of
the type and structure of it's argument in order to understand that its
second parameter has to be passed as an argument to its first
parameter. But, that makes them strongly coupled right? What if I later
wanted to use Foo with a function that was not a template?

Anyway, this is interesting to know.

Thanks for your help ... I finally went with Functors anyway!

regards,

-vijai.

Nov 8 '05 #5

This discussion thread is closed

Replies have been disabled for this discussion.