std::function extract and remove argument

Question

I'm learning c++11/14 these days and it seems like a whole new language to me with all the great additions but I still can't quite make use of all these new features:

typedef std::function<void()> void_f;

typedef std::function<void(int a, void_f b)> f1;
typedef std::function<void(int a, std::function<void(void_f f)> c, void_f b)> f2; // this order is wanted

std::vector<f1> v;

void add(f1 f)
{
    v.push_back(f);
}

void add(f2 f)
{
     v.push_back(f) // ?
 // I want to extract 'void_f f' argument from std::function for later use 
//and remove std::function completely to get same signature as f1
}

I've been looking at std::move, std::bind, std::forward, std::placeholders but I can't seem to find anything like this. Maybe I should save longer version in vector and then bind empty lambda for shorter one?

what do you mean by *"I want to extract 'void_f f' argument from std::function for later use and remove std::function completely"* ? — Piotr Skotnicki, Aug 11 '15 at 19:50
A `std::function` is something that can be called with its arguments, not something that *has* arguments? — Yakk - Adam Nevraumont, Aug 11 '15 at 19:52
**Why** do you want to extract the function pointer from the `std::function`? What are you hoping that will offer you? *Note that there may be no function pointer at all* — Drew Dormann, Aug 11 '15 at 19:56
I've updated my question to reflect better what I want to achieve. I just want to save in the same vector holder, but also reuse second argument from std::function — Petar, Aug 11 '15 at 20:28
What do you want to *reuse* ? `f` itself doesn't hold any argument to be *reused* — Piotr Skotnicki, Aug 11 '15 at 20:33
f is a void function that can have some code that can be called later on — Petar, Aug 11 '15 at 20:54
the second and third typedef's don't make any sense and don't compile — bolov, Aug 11 '15 at 21:11

score 1 · Accepted Answer · answered Aug 12 '15 at 08:10

1

No, you can save the "shorter" f1 in the vector if that's what you want:

typedef std::function<void()> void_f;

typedef std::function<void(int a, void_f b)> f1;
typedef std::function<void(int a, std::function<void(void_f f)> c, void_f b)> f2;

std::vector<f1> v;

void add(f1 f)
{
    v.push_back(f);
}

void add(f2 f)
{
     v.push_back([](int a, void_f b) { f(a, [](){}, b); });
}

You can't "remove" the extra argument of f2 but you can pass a no-op lambda. The outer lambda puts the remaining arguments a and b into the right order.

answered Aug 12 '15 at 08:10

MSalters

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that was what I've done first but then argument is completely lost. Is there any way to get ahold of it somehow? – Petar Aug 12 '15 at 08:50
@juzerKicker: Which argument is lost? The whole reason why we provide an empty lambda as a second argument is because there wasn't an argument in the first place. `f2 f` needs 3 arguments, and the users of `vector` will only provide 2 arguments. – MSalters Aug 12 '15 at 12:38
std::function has f argument, which is lambda, which may contain code written by hand, usage would be like: c([]{} _some_code_here}). I would like to save that before 'removing' it by calling no-op lambda later on. I think I will do that on the fly once c is called. – Petar Aug 12 '15 at 13:39
1

@juzerKicker: Higher-order functions can be confusing. `std::function` doesn't have an `f` argument, it _needs_ an argument. – MSalters Aug 12 '15 at 18:57

std::function extract and remove argument

1 Answers1