CODEWITHSUNDEEP
c++
Riptide
Riptide

Reputation: 405

can a lambda be inherent to a class

Would it be possible for a lambda, which is inside a class X and defined outside that class X to magically use member variables of that class X ?

Like so:

struct Foo {
    int x;
    std::function<void(void)> f;
};

int main() {
    Foo foo;

    foo.x = 10;
    foo.f = []() {
        std::cout << x;
    };

    foo.f();
}

It would magically, without any context, use the "x" from the Foo struct.

I thought that because of a lambda being an anonymous struct, if I pass the object "foo" (the C way) in that function, it works, obviously:

struct Foo {
    int x;
    std::function<void(Foo&)> f;
};

int main() {
    Foo foo;

    foo.x = 10;
    foo.f = [](Foo& me) {
        std::cout << me.x;
    };

    foo.f(foo);
}

Is there any workaround to this ?

Upvotes: 1

Views: 55

Answers (2)

Raven
Raven

Reputation: 345

without any context, you can do this

foo.f = [x=foo.x]() {
   std::cout << x;
};

Upvotes: 2

Kerrek SB
Kerrek SB

Reputation: 476960

The natural approach would be to use a constructor:

struct Foo {
    Foo() : f([this](){ std::cout << x; }) {}

    int x;
    std::function<void()> f;
};

Or perhaps a default member initializer:

struct Foo {
    int x;
    std::function<void()> f = [this]() { std::cout << x; };
};

(The function doesn't take any arguments, of course, so I changed the type of Foo::f accordingly.)

The lambda cannot be "inherent to a class", since the lambda refers to an object of the class type, so every object must get its own lambda.

Usage:

Foo m;
m.x = 5;
m.f();

Upvotes: 3

Related Questions