why this std::bind fails? - c++

Question

What is the fault of this bind operation with visual studio 2013 for error C3867 ?
#include <map>
#include <vector>
#include <algorithm>
#include <functional>
int main()
{
std::map<int, int> m1, m2;
std::vector<std::map<int, int> *> pM;
std::for_each(pM.begin(), pM.end(),
std::bind(std::map<int, int>::erase , 1));
}

Answer 1

If you want to delete each map's element (which key is 1) in the vector. Here is a sample.
#include <map>
#include <vector>
#include <algorithm>
#include <functional>
#include <iostream>
using namespace std;
using namespace std::placeholders;
int main()
{
std::map<int, int> m1 {{1,2},{2,3},{3,4}}, m2 {{2,3},{1,5},{8,9}};
std::vector<std::map<int, int> *> pM {&m1,&m2};
size_t(std::map<int,int>::*pf)(const int& key)=&std::map<int,int>::erase;//note this line.
std::for_each(pM.begin(), pM.end(),std::bind(pf,_1,1));//note this line
for(auto& ele:m1)
{
cout<<ele.first<<","<<ele.second<<endl;
}
for(auto& ele:m2)
{
cout<<ele.first<<","<<ele.second<<endl;
}
}
As chris says:
For one thing, member functions do not decay into member function
pointers. For another, std::map::erase is overloaded.
We must use & to get member function pointer,and must tell compiler which overload function you want to choice.
So write code like this:
size_t(std::map<int,int>::*pf)(const int& key)=&std::map<int,int>::erase;
And further：
std::bind(pf,_1,1)
member function need object as its implicit first parameter, _1 do this work.The last parameter 1 is int as the key pass to map erase member function.