I am trying to learn std::function
and here's my code:
#include <iostream>
#include <functional>
struct Foo {
void print_add(int i){
std::cout << i << '\n';
}
};
typedef std::function<void(int)> fp;
void test(fp my_func)
{
my_func(5);
}
int main(){
Foo foo;
test(foo.print_add);
return 0;
}
Compiler Error:
error: cannot convert 'Foo::print_add' from type 'void (Foo::)(int)' to type 'fp {aka std::function<void(int)>}'
test(foo.print_add);
How can I make this work, i.e how can I pass a member function as a parameter?
print_add
is a non-static member function of foo
, which means it must be invoked on an instance of Foo
; hence it has an implicit first argument, the this
pointer.
Use a lambda that captures the foo
instance and invokes print_add
on it.
Foo foo;
test([&foo](int i){ foo.print_add(i); });
Another option is to use std::bind
to bind the foo
instance:
test(std::bind(&Foo::print_add, &foo, std::placeholders::_1));
Live demo
You need an object to call a non-static member function. Hence, when you want to get a std::function
you have several options:
std::function
std::function
is called#include <iostream>
#include <functional>
struct A {
int i = 42;
int get() const { return i; }
static int get_static() { return 0; }
};
struct A_functor {
A a;
int operator()() const { return a.get(); }
};
int main() {
// static method
std::function<int()> f1 = &A::get_static;
std::cout << f1() << "\n";
// bind an object to the function via lambda
A a;
std::function<int()> f2 = [&a](){ return a.get(); };
std::cout << f2() << "\n";
// store the object in the std::function
std::function<int()> f3 = A_functor{};
std::cout << f3() << "\n";
// or
std::function<int()> f4 = [a = A()](){ return a.get(); };
std::cout << f4() << "\n";
// pass the object when the std::function is called
std::function<int(A&)> f5 = &A::get;
std::cout << f5(a) << "\n";
// or
std::function<int(A*)> f6 = &A::get;
std::cout << f6(&a) << "\n";
}
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