Allowing a functor or a function to be stored in a class

[James Aguilar]

In the C++ STL, there are various places where you can call a function that
takes a functor as a parameter. This functor may be either a function object or
a function pointer. For instance, if you have a vector<char> and call for_each
on it, both of the methods below would be valid:

#include <iostream>
#include <vector>
using namespace std;

struct MyFunctionObject
{
void operator ()(char a) {cout << a;};
};
void myFunction(char a) {cout << a;}

int main()
{
vector<char> v;
//fill it . . .
for_each(v.begin(), v.end(), (MyFunctionObject()));
for_each(v.begin(), v.end(), &myFunction);
return 0;
}

My question pertains to a similar issue: how do I cause the same to occur in a
class template? I would like my template to look like this:

template <typename CompT>
class WhyWontThisWork
{
CompT m_comparator;
public:
WhyWontThisWork(CompT cmpIn) : m_comparator(cmpIn) {}
void doThis(char a, char b) { if (m_comparator(a, b)) cout << "WIN"; }
};

Can I actually do this in C++ without writing an inordinate amount of code? If
so, how?

- JFA1

 

[Pete Becker]

My question pertains to a similar issue: how do I cause the same to occur in a
class template? I would like my template to look like this:

template <typename CompT>
class WhyWontThisWork
{
CompT m_comparator;
public:
WhyWontThisWork(CompT cmpIn) : m_comparator(cmpIn) {}
void doThis(char a, char b) { if (m_comparator(a, b)) cout << "WIN"; }
};

Can I actually do this in C++ without writing an inordinate amount of code? If
so, how?

This works just fine. What problem did you encounter?

struct Comp
{
bool operator()(char a, char b) const { return a == b; }
};

int main()
{
Comp comp;
WhyWontThisWork<Comp> cmp(comp);
cmp.doThis('a', 'b');
cmp.doThis('a', 'a');
return 0;
}

 

[Victor Bazarov]

In the C++ STL, there are various places where you can call a function that
takes a functor as a parameter. This functor may be either a function object or
a function pointer. For instance, if you have a vector<char> and call for_each
on it, both of the methods below would be valid:

#include <iostream>
#include <vector>
using namespace std;

struct MyFunctionObject
{
void operator ()(char a) {cout << a;};
};
void myFunction(char a) {cout << a;}

int main()
{
vector<char> v;
//fill it . . .
for_each(v.begin(), v.end(), (MyFunctionObject()));
for_each(v.begin(), v.end(), &myFunction);
return 0;
}

My question pertains to a similar issue: how do I cause the same to occur in a
class template? I would like my template to look like this:

template <typename CompT>
class WhyWontThisWork
{
CompT m_comparator;
public:
WhyWontThisWork(CompT cmpIn) : m_comparator(cmpIn) {}

I don't understand 'WhyWontThisWork' bit of it.

void doThis(char a, char b) { if (m_comparator(a, b)) cout << "WIN"; }
};

Can I actually do this in C++ without writing an inordinate amount of code?

What's inordinate?

I don't see you even attempting to use your template. How can we continue
the discussion without you even trying?

If
so, how?

What book are you reading that doesn't explain templates? What section of
the FAQ regarding questions about some code that doesn't work don't you
understand? Am I using overly long sentences?

V

 

[James Aguilar]

I don't see you even attempting to use your template. How can we continue
the discussion without you even trying?

OK.

//=== CODE ===
#include <iostream>

using namespace std;

template <typename CompT>
class Test
{
CompT m_comparator;
public:
Test(CompT cmpIn) : m_comparator(cmpIn) {}
void doThis(char a, char b) { if (m_comparator(a, b)) cout << "WIN"; }
};

class TestFunctionObject
{
public:
bool operator ()(char a, char b) {return a < b;}
};

bool testFunction(char a, char b)
{
return a < b;
}

int main()
{
Test a(&testFunction);
Test b((TestFunctionObject()));

a.doThis('a', 'b');
b.doThis('a', 'b');

return 0;
}
//=== CODE ===

For obvious reasons, this code does not compile. I didn't provide template
types for a or b in main. But that is really the question I am asking: is it
possible for me to have a class that can take a function pointer or a function
object in the constructor and treat them the same way (i.e. store them for later
use in another function)? If so, what is the type I need to use to replace
CompT?

- JFA1

 

[Pete Becker]

Test(CompT cmpIn) : m_comparator(cmpIn) {}
If so, what is the type I need to use to replace
CompT?

Since CompT is the type of the argument that you're going to pass to the
constructor, that's the type you should specify. That's also why the
standard library provides the template functions bind1st() and bind2nd()
in addition to the template classes binder1st and binder2nd.

 

[Victor Bazarov]

OK.

//=== CODE ===
#include <iostream>

using namespace std;

template <typename CompT>
class Test
{
CompT m_comparator;
public:
Test(CompT cmpIn) : m_comparator(cmpIn) {}
void doThis(char a, char b) { if (m_comparator(a, b)) cout << "WIN"; }
};

class TestFunctionObject
{
public:
bool operator ()(char a, char b) {return a < b;}
};

bool testFunction(char a, char b)
{
return a < b;
}

int main()
{
Test a(&testFunction);
Test b((TestFunctionObject()));

a.doThis('a', 'b');
b.doThis('a', 'b');

return 0;
}
//=== CODE ===

For obvious reasons, this code does not compile.

Well, DUH!

I didn't provide template
types for a or b in main. But that is really the question I am asking: is it
possible for me to have a class that can take a function pointer or a function
object in the constructor and treat them the same way (i.e. store them for later
use in another function)? If so, what is the type I need to use to replace
CompT?

Of course not. "Test" is a template. It needs an argument. The class
'TestFunctionObject' and the function 'testFunction' have different types.
You cannot expect your 'Test' template to become one and the same if you
instantiate it with two different types, now, can you?

You can always make your 'Test' class take a functor but to use, say,
functions with it you'd have to provide a separate object. Example:

struct myFunctor_base {
virtual void operator() const (char a, char b) = 0;
};

struct function_to_myFunctor_adapter : myFunctor_base {
void (*pfunc)(char,char);
function_to_myFunctor_adapter(void (*p)(char,char)) : pfunc(p) {}
void operator()(char a, char b) const {
return pfunc(a,b);
}
};

struct myFunctor : myFunctor_base {
void operator()(char,char) const; // implemented elsewhere
};

class Test {
myFunctor_base const& f;
public:
Test(myFunctor_base const& ff) : f(ff) {}
void doThis(char a, char b) { f(a,b); }
};

void myFunction(char a, char b);

int main() {
Test a = Test(myFunctor());
Test b = Test(function_to_myFunctor_adapter(myFunction));
a.doThis('a', 'b');
b.doThis('a', 'b');
}

Now, this probably constitutes (how did you put it?) inordinate amount
of code, but that's essentially how you can do it.

V

 

[James Aguilar]

Now, this probably constitutes (how did you put it?) inordinate amount
of code, but that's essentially how you can do it.

That's exactly how I put it, and yeah, it meets the criteria. I appreciate your
writing that code -- I probably wouldn't have if someone was asking me the same
question.

I take it that the solution to my problem is thus to choose one or the other
(function pointers or function objects) and stick with it, and not try to make
my stuff compatible with both?

- JFA1

 

[Pete Becker]

You can always make your 'Test' class take a functor but to use, say,
functions with it you'd have to provide a separate object.

It works as is, but you have to get the type of the template argument right:

Test<bool (*const)(char,char)> wrap(testFunction);

The reason that the standard library templates need wrapper classes for
function pointers is to provide the nested typedefs result_type and the
various argument_types. The TR1 function adaptors don't need these
typedefs, so, like Test, they can be used with raw function pointers.
(See my article in the upcoming May issue of CUJ).

 

[Pete Becker]

Test a = Test(myFunctor());
Test b = Test(function_to_myFunctor_adapter(myFunction));

Oh, I see, you're solving a different problem from the one I was
solving. The original template can be instantiated with various types;
if you need a single type that can hold different types of target
objects you need TR1's template class "function":

function<bool(char,char)> fp;
fp = TestFunctionObject();
fp = testFunction;

(Okay, you have to add "typedef bool result_type" to TestFunctionObject
to get this to compile with normal C++ compilers)

 

[Victor Bazarov]

Oh, I see, you're solving a different problem from the one I was solving.
The original template can be instantiated with various types; if you need
a single type that can hold different types of target objects you need
TR1's template class "function":

function<bool(char,char)> fp;
fp = TestFunctionObject();
fp = testFunction;

This is neat. I didn't know that. How many compilers support this?
Is 'function' template class part of 'std'? Which header to include?

(Okay, you have to add "typedef bool result_type" to TestFunctionObject to
get this to compile with normal C++ compilers)

"Normal"? What do you mean? Is it already supported?

Thanks.

V

 

[Pete Becker]

This is neat. I didn't know that. How many compilers support this?
Is 'function' template class part of 'std'? Which header to include?

It's part of TR1, not generally available yet (TR1 has just gone out for
the first of two ballots). Boost has function's predecessor, which may
have been updated to match the TR1 spec.

"Normal"? What do you mean? Is it already supported?

Sorry, I was being too clever. The problem is that standard C++ code
can't extract the return type of a function object, so implementations
of function (and some of the other call wrappers in TR1) need a little
help. There's a template in TR1 that figures out return types, and with
some compiler help it can get the type completely right. Without that
compiler help, function objects have to define result_type. They don't
have to have the various argument_type typedefs, though. So
implementations of TR1 for "normal" compilers (i.e. compilers that don't
have a special hook for getting that return type) will need "result_type".

 

[Pete Becker]

It's part of TR1, not generally available yet (TR1 has just gone out for
the first of two ballots). Boost has function's predecessor, which may
have been updated to match the TR1 spec.

I forgot to mention where to find the latest draft of TR1:
www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1745.pdf