I am trying to create a Functor using boost function and bind, but i cant pass only a single argument to the destination function which has 3 arguments:
#include <boost/bind.hpp>
#include <boost/function.hpp>
template <typename Functor>
void foreach(Functor f)
{
int k;
f(k);
}
class A
{
private:
void bar(int &a,int &b,int& c)
{}
void foo()
{
int keys,predicate;
boost::function<void(int &,int&,int&)> functor_ ( boost::bind(
&A::bar,
this,
boost::ref(keys),
boost::ref(predicate),
_1
));
foreach(functor_);
}
};
The error says:
binf.cpp: In instantiation of ‘void foreach(Functor) [with Functor = boost::function<void(int&, int&, int&)>]’:
binf.cpp:29:18: required from here
binf.cpp:9:6: error: no match for call to ‘(boost::function<void(int&, int&, int&)>) (int&)’
f(k);
^
In file included from /usr/include/boost/function/detail/maybe_include.hpp:28:0,
from /usr/include/boost/function/detail/function_iterate.hpp:14,
from /usr/include/boost/preprocessor/iteration/detail/iter/forward1.hpp:62,
from /usr/include/boost/function.hpp:64,
from binf.cpp:2:
/usr/include/boost/function/function_template.hpp:1048:7: note: candidate is:
class function<BOOST_FUNCTION_PARTIAL_SPEC>
^
/usr/include/boost/function/function_template.hpp:761:17: note: boost::function3<R, T1, T2, T3>::result_type boost::function3<R, T1, T2, T3>::operator()(T0, T1, T2) const [with R = void; T0 = int&; T1 = int&; T2 = int&; boost::function3<R, T1, T2, T3>::result_type = void]
result_type operator()(BOOST_FUNCTION_PARMS) const
^
/usr/include/boost/function/function_template.hpp:761:17: note: candidate expects 3 arguments, 1 provided
Of course, I want to pass only one argument, because I expect bind to take care of the other 2 args. I tried to search for my error but I couldn't find any(in a short call). Can you please help me find the problem? Thanks
You are creating boost::function<void(int&, int&, int&)>
, i.e. function, that takes three arguments. Since you are using bind and want to call function with one argument it should be just boost::function<void(int&)>
.