compiling the following code via
g++ -std=c++11 test.cpp
gives me the following error:
test.cpp: In lambda function:
test.cpp:17:128: error: passing ‘const CMyclass’ as ‘this’ argument of ‘void CMyclass::my_method(const state_type&, double)’ discards qualifiers [-fpermissive]
  std::function<void(const state_type &,const double)> observer = [=](const state_type &x,const double t){my_class.my_method(x,t);};
                                                                                                                                ^
I have searched other similar questions but I can't figure out what is wrong with this code.
#include <vector>
#include <functional>
typedef std::vector<int> state_type;
class CMyclass
{
public:
    void my_method( const state_type &x , const double t )
    {
    }
};
int main()
{
    CMyclass my_class;
    std::function<void(const state_type &,const double)> observer =
         [=](const state_type &x,const double t)
         {
             my_class.my_method(x,t);
         };
}
edit:
I would not put const after the method.
Since you're taking by value my_class becomes const-qualified. You have three options to fix it:
Add const to your method:
void my_method(const state_type &x, const double t) const
Capture by reference:
[&](const state_type &x, const double t) { .. }
Or make the lambda mutable:
[=](const state_type &x,const double t) mutable { .. }
You have to mark your lambda mutable
observer = [=](const state_type &x,const double t) mutable {my_class.my_method(x,t);};
full code below
#include <vector>
#include <functional>
typedef std::vector<int> state_type;
class CMyclass
{
public:
    void my_method( const state_type &x , const double t )
    {
    }
};
int main()
{
    CMyclass my_class;
    std::function<void(const state_type &,const double)> observer = 
        [=](const state_type &x,const double t) mutable {
            my_class.my_method(x,t);
        };
}
compiled by g++ -Wall -std=c++11, gcc v4.9.2
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