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

Discussion in 'C++' started by James Aguilar, Mar 29, 2005.

  1. 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
    James Aguilar, Mar 29, 2005
    #1
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  2. James Aguilar

    Pete Becker Guest

    James Aguilar wrote:
    >
    > 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;
    }

    --

    Pete Becker
    Dinkumware, Ltd. (http://www.dinkumware.com)
    Pete Becker, Mar 29, 2005
    #2
    1. Advertising

  3. James Aguilar wrote:
    > 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
    Victor Bazarov, Mar 29, 2005
    #3
  4. "Victor Bazarov" <> wrote in message
    news:IPk2e.58298$01.us.to.verio.net...
    >
    > 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
    James Aguilar, Mar 29, 2005
    #4
  5. James Aguilar

    Pete Becker Guest

    James Aguilar wrote:
    > 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.

    --

    Pete Becker
    Dinkumware, Ltd. (http://www.dinkumware.com)
    Pete Becker, Mar 30, 2005
    #5
  6. James Aguilar wrote:
    > "Victor Bazarov" <> wrote in message
    > news:IPk2e.58298$01.us.to.verio.net...
    >
    >>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.


    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
    Victor Bazarov, Mar 30, 2005
    #6
  7. "Victor Bazarov" <> wrote in message
    news:tql2e.58303$01.us.to.verio.net...
    >
    > 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
    James Aguilar, Mar 30, 2005
    #7
  8. James Aguilar

    Pete Becker Guest

    Victor Bazarov wrote:
    >>
    >> 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"; }
    >> };
    >>
    >>
    >> bool testFunction(char a, char b)
    >> {
    >> return a < b;
    >> }
    >>

    >
    > 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
    Dinkumware, Ltd. (http://www.dinkumware.com)
    Pete Becker, Mar 30, 2005
    #8
  9. James Aguilar

    Pete Becker Guest

    Victor Bazarov wrote:
    >
    > 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)

    --

    Pete Becker
    Dinkumware, Ltd. (http://www.dinkumware.com)
    Pete Becker, Mar 30, 2005
    #9
  10. "Pete Becker" <> wrote...
    > Victor Bazarov wrote:
    >>
    >> 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;


    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
    Victor Bazarov, Mar 30, 2005
    #10
  11. James Aguilar

    Pete Becker Guest

    Victor Bazarov wrote:
    >
    > 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.

    >
    >>(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?
    >


    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
    Dinkumware, Ltd. (http://www.dinkumware.com)
    Pete Becker, Mar 30, 2005
    #11
  12. James Aguilar

    Pete Becker Guest

    Pete Becker, Mar 30, 2005
    #12
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