passing ‘const CMyclass’ as ‘this’ argument of ... discards qualifiers [-fpermissive]

Question

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.

score 22 · Accepted Answer · answered Dec 20 '14 at 20:43

22

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 { .. }

answered Dec 20 '14 at 20:43

David G

94,763
41
167
253

second option was exactly what I needed. – barej Dec 20 '14 at 20:46

score 1 · Answer 2 · edited Dec 20 '14 at 20:49

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

score 0 · Answer 3 · answered Dec 20 '14 at 20:43

passing const CMyclass as this argument of void CMyclass::my_method(const state_type&, double) discards qualifiers

The error message says all. void CMyclass::my_method(const state_type&, double) expects a normal instance but not a const one. Change the method definition to

void CMyclass::my_method(const state_type&, double) const;

to resolve the error, or change the [=] lambda initialiser to whatever you need there (e.g. [&], but make sure it is not const.

passing ‘const CMyclass’ as ‘this’ argument of ... discards qualifiers [-fpermissive]

3 Answers3