boost bind class function pointer

Question

class Foo 
{
    double f1( int x, std::string s1 );
    double f2( int x, SomeClass s2 );
}

I want to be able to bind Foo.f1's s1 without an instance of foo to create in essense

typedef double (Foo::* MyFooFunc)( int ) 

MyFooFunc func1 = boost::bind( &Foo::f1, _1, _2, "some string" );
MyFooFunc func2 = boost::bind( &Foo::f2, _1, _2, SomeClass );

Then I pass func1 and func2 as parameters to other functions, inside which Foo is finally bound:

void SomeOtherFunction( MyFooFunc func )
{
     Foo foo;
     boost::function< double (int) > finalFunc =
          boost::bind( func, foo, _1 );
}

Questions: Is this possible? If yes, 1) how to achieve it? 2) What's the declaration of MyFooFunc?

It depends on SomeOtherFunction is declared. Note that your second typedef is not a function pointer, it's pointer-to-member-function (PTMF), which is an entirely different and incompatible concept. Wrapping everything in std::function or std::tr1::function or boost::function would probably be the most powerful solution. — Kerrek SB
– Kerrek SB, Commented Aug 5, 2011 at 14:31
There are actually two issues here: 1) binding PTMF without an instance. 2) express the result of the bind as boost function. For 1), I tried boost::bind( &Foo::f1, _1, _2, "hello" ). That didn't work. For 2), I suspect it's boost::function< int (boost::shared_ptr<Foo>,int) >. — Candy Chiu
– Candy Chiu, Commented Aug 5, 2011 at 14:53

Luc Danton · Accepted Answer · 2011-08-05 15:33:31Z

typedef double (Foo::* MyFooFunc)( int );

MyFooFunc func1 = boost::bind( &Foo::f1, _1, _2, "some string" );

The result of boost::bind is not a pointer to member, so func1 cannot be initialized as such on the second line. The result of boost::bind is an unspecified type (which will depend on the parameters). If you're using C++0x, the simplest way to name the result of a call to bind is to use auto:

auto func1 = boost::bind( &Foo::f1, _1, _2, "some string" );

Another simple way (not restricted to C++03) is simply to not name the result, but to use it on the spot:

SomeOtherFunction(boost::bind(&Foo::f1, _1, _2, "some string"));

Or, you can use type-erasure to store the result of boost::bind into a boost::function, which you seem to be familiar with. boost::function<double(Foo&, int)> is a possibility but not the only choice.

We now need to find the appropriate signature for SomeOtherFunction: again, a pointer to member can't be initialized from the result of a call to boost::bind, so void SomeOtherFunction(MyFooFunc func); won't work. You can make the function a template instead:

template<typename Func>
void SomeOtherFunction( Func func )
{
     Foo foo;
     boost::function< double (int) > finalFunc =
          boost::bind( func, foo, _1 );
}

If a template is not preferrable, then you must use some kind of type-erasure such as, again, boost::function.

void SomeOtherFunction(boost::function<double(Foo&, int)> const& func);

(once again other boost::function types are possible depending on details such as passing a ref-to-const as opposed to a ref-to-non-const)

Jens Luedicke · Accepted Answer · 2011-08-05 14:32:26Z

0

Try this:

boost::bind(&Foo::f1, object, _1, _2);

object is an instance of class Foo. _1 and _2 are the argument placeholders.

answered Aug 5, 2011 at 14:32

Jens Luedicke

1,0042 gold badges9 silver badges20 bronze badges

3 Comments

Kerrek SB Over a year ago

There's probably just one placeholder, since the OP wants Foo::f1(int) I think...

Cat Plus Plus Over a year ago

Note that if object is a value and not a pointer, you should do boost::ref(object) or &object, otherwise binder will operate on a copy.

Candy Chiu Over a year ago

Sorry for the confusion. I want to separate the binding of an instance of Foo (object) and the non uniform parameters (s1,s2) into 2 steps.

Collectives™ on Stack Overflow

boost bind class function pointer

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related