This is the answer of Exercise #9: No for-loops.
Replace the for-loop. You will need:
#include <vector> struct Widget { void DoItOften(const int n) const { /* do it n times */ } }; void DoItOften(const std::vector<Widget>& v, const int n) { const int sz = v.size(); for (int i=0; i!=sz; ++i) { v[i].DoItOften(n); } }
Answer using STL only
#include <algorithm> #include <numeric> #include <vector> struct Widget { void DoItOften(const int n) const { /* do it n times */ } }; void DoItOften(const std::vector<Widget>& v, const int n) { std::for_each( v.begin(), v.end(), std::bind2nd(std::mem_fun_ref(&Widget::DoItOften),n)); }
Answer using Boost
#include <algorithm> #include <vector> #include <boost/bind.hpp> struct Widget { void DoItOften(const int n) const { /* do it n times */ } }; void DoItOften(const std::vector<Widget>& v, const int n) { std::for_each( v.begin(), v.end(), boost::bind(&Widget::DoItOften, _1, n)); }
Note that you do not need boost::mem_fn, because it is added for you. If this is done by hand, like in the code below, the solution is still correct.
void DoItOften(const std::vector<Widget>& v, const int n) { std::for_each( v.begin(), v.end(), boost::bind(boost::mem_fn(&Widget::DoItOften), _1, n)); }
Note that _1 is a placeholder of type boost::arg<1> and can be found in boost/bind/placeholders.hpp.