Can I generate a function without providing arguments?

问题

So c++17 has std::function Deduction Guides so given:

int foo();

I can do:

std::function bar(foo);

But I'm stuck on a c++14 compiler. There I have to do something more like: function<int()> bar(foo). I was wondering if there was a way to create a std::function without passing the function pointer and explicitly providing the function signature? So for example make_pair will deduce the type of it's return from it's arguments. I was wondering if I could write something similar for functions even using c++14, like:

auto bar = make_function(foo);

Is this doable?

_{Note: My real case is that foo is a template function with a lot of arguments I don't want to deduce. So my motivation here is to generate a function without needing to provide the parameter types.}

Live Example

回答1:

Your question has some most important part in the end in the fine print. If your foo is a template, C++17 deduction guides won't help you with a simple syntax like

std::function f(foo);

You'd still need to provide template arguments for foo. Assuming you are OK with specifying foo's argument types (as you have to be) writing make_func is a trivial exercise:

 template<class R, class... ARGS>
 auto make_func(R (*ptr)(ARGS...)) {
      return std::function<R (*)(ARGS...)>(ptr);
 }

And than you use it:

auto bar = make_func(&foo<Z, Y, Z>);

来源：https://stackoverflow.com/questions/54990207/can-i-generate-a-function-without-providing-arguments