c++ lambda expressions variables in classes - c++

Question

I want to save lambda expressions variables (like in the fist code block). The problem is that then I use a classes (like the second code block) the compiler return me some errors. I don"t know how to fix it.
I hope somebody can help me and explain, why it's not working like this. Thanks.
First Code:
// variable for function pointer
void (*func)(int);
// default output function
void my_default(int x) {
cout << "x =" << "\t" << x << endl << endl;
}
int main() {
cout << "Test Programm\n\n";
// 1. Test - default output function
cout << "my_default\n";
func = &my_default;
func(5);
// 2. Test - special output function 2
cout << "my_func2\n";
func = [](int x) { cout << "x =" << " " << x << endl << endl; };
func(5);
return 0;
}
Second Code:
class test {
private:
// variable for function pointer
void (*func)(int);
// default output function
void my_default(int x) {
cout << "x =" << "\t" << x << endl << endl;
}
public:
void dummy(void) {
// 1. Test - default output function
cout << "my_default\n";
func = &my_default;
func(5);
// 2. Test - special output function 2
cout << "my_func2\n";
func = [](int x)->int{ cout << "x =" << " " << x << endl << endl; };
func(5);
}
};
// entry
int main() {
cout << "Test Programm\n\n";
test a;
a.dummy();
return 0;
}
Compiler:
pi#raspberrypi ~/dev/property $ gcc -std=c++0x -o test2 test2.cpp -lstdc++
test2.cpp: In member function ‘void test::dummy()’:
test2.cpp:491:17: error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say ‘&test::my_default’ [-fpermissive]
test2.cpp:491:17: error: cannot convert ‘void (test::*)(int)’ to ‘void (*)(int)’ in assignment
test2.cpp:496:77: error: invalid user-defined conversion from ‘test::dummy()::<lambda(int)>’ to ‘void (*)(int)’ [-fpermissive]
test2.cpp:496:28: note: candidate is: test::dummy()::<lambda(int)>::operator int (*)(int)() const <near match>
test2.cpp:496:28: note: no known conversion for implicit ‘this’ parameter from ‘int (*)(int)’ to ‘void (*)(int)’

Answer 1

The problem is that member functions are not normal functions, they can't be assigned to a function pointer because the type in which they are member is part of their type.
Also, a member function needs to have an object to be called on, which will be the this inside the function code.
You have several solutions:
allow only functions which are member of your class
void (*MyClass::func)(int); // but you can use it only with members of the class
use std::function
typedef std::function<void(int)> func;
The solution 2 is the simplest as std::function is designed to work with anything that is callable with the same signature as the one in the template parammetters.
Also, it's the only solution that allows you to store closures(objects from lambdas).
See C++11 styled callbacks? for details.
class test {
private:
// variable for function pointer
std::function< void ( int )> func;
// default output function
void my_default(int x) {
cout << "x =" << "\t" << x << endl << endl;
}
public:
void dummy(void) {
// 1. Test - default output function
cout << "my_default\n";
func = std::bind(&test::my_default, this, std::placeholders::_1);
// or
func = [&]( int i ){ my_default( i ); };
func(5);
// 2. Test - special output function 2
cout << "my_func2\n";
func = [](int x)->int{ cout << "x =" << " " << x << endl << endl; };
func(5);
}
};
// entry
int main() {
cout << "Test Programm\n\n";
test a;
a.dummy();
return 0;
}

Answer 2

A member function is not like an ordinary function, in that there has to be an instance of the class available in order to invoke it (i.e., the object that will become *this). You can't make an ordinary function pointer variable point to a member function.
If you want to create a function pointer that can be called using any instance of the class, you need a member function pointer. You would write
void (test::*func)(int);
to declare it,
func = &test::my_default;
to assign it, and
(this->*func)(5);
to call it. Of course, now you can't make a member function pointer point to a lambda.
If on the other hand you want to bind this as the instance and create an ordinary function from a member function, well, you can't actually make an ordinary function pointer to it. Instead you'll want an std::function object,
std::function<void(int)> func;
bind as follows:
func = std::bind(&test::my_default, this, std::placeholders::_1);
and then call normally. std::function works with lambdas just like a function pointer would.