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?
Try this:
object
is an instance of class Foo. _1 and _2 are the argument placeholders.The result of
boost::bind
is not a pointer to member, sofunc1
cannot be initialized as such on the second line. The result ofboost::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 tobind
is to useauto
:Another simple way (not restricted to C++03) is simply to not name the result, but to use it on the spot:
Or, you can use type-erasure to store the result of
boost::bind
into aboost::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 toboost::bind
, sovoid SomeOtherFunction(MyFooFunc func);
won't work. You can make the function a template instead:If a template is not preferrable, then you must use some kind of type-erasure such as, again,
boost::function
.(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)