Let\'s say, I have a class:
class A {
int a;
};
And I have a lambda:
auto function = [](A* a) {
a->a; // <== giv
You can do it by creating a friend function that returns the lambda function. It inherits the friend access:
struct A {
friend std::function<void(A&, int)> f();
private:
int i;
void test() {std::cout << "test: " << i << "\n";}
};
std::function<void(A&, int)> f() {
return [] (A &a, int i) {a.i = i; a.test(); };
}
int main() {
A a;
f()(a, 13);
return 0;
}
using std::function takes extra resource, so I recomendet using friend/or method function to access private member (friend function implicit inlined):
class A{
int a;
friend int access_member(A*a){ return a->a;}
};
-----------------------------------------
auto function = [](A*a){ return access_member(a); }
Live example
EDIT: I personally like std::function, but don't forgot, std::function always takes extra memory resources, and may not inlined , so if you may implement your source without std::function, don't use std::function. See, How is std::function implemented? Also, Lambda to std::function conversion performance
In order to make a lambda a friend, you need to befriend a class or a function where the lambda is defined. Here is a complete example:
#include <iostream>
using namespace std;
class A {
int a;
public:
A(int _a) : a(_a) {}
friend int foo(A*); // Declare foo(A*) a friend of A
};
int foo(A* aa) {
auto function = [](A* a) {
return a->a; // Now foo(A*) can access A::a, which is private
};
return function(aa);
}
int main() {
A a(123);
cout << foo(&a) << endl;
return 0;
}
Here is a running demo on ideone.