Class's operator() or bind a function as a functor?

Class's operator() or bind a function as a functor?

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Class's operator() or bind a function as a functor?
Tag : performance , By : quicky
Date : November 25 2020, 01:01 AM

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How do you write a operator() or less-than-functor neater than a trivalue-compare-function

Tag : cpp , By : Killercode
Date : March 29 2020, 07:55 AM
hope this fix your issue The thing is that you are fine with just declaring one trivalue compare function if you autogenerate all operators using: http://en.wikipedia.org/wiki/Barton%E2%80%93Nackman_trick

Why functor result bind to new module name is necessary for call nested functor of?

Tag : ocaml , By : Sanoran
Date : March 29 2020, 07:55 AM
this will help While I don't believe that this restriction is strictly necessary, it is probably motivated by certain limitations in OCaml's module type system. Without going into too much technical detail, OCaml requires all intermediate module types to be expressible as syntactic signatures. But with functors, that sometimes isn't possible. For example, consider:
module Functor(X : sig end) = struct
  type t = T of int
  module Nested(Y : sig end) = struct let x = T 5 end 
functor(Y : sig end) -> sig val x : Functor(struct end).t end  (* not legal OCaml! *)
module A = Functor(struct end)
functor(Y : sig end) -> sig val x : A.t end

Using operator new as a functor/function pointer

Tag : cpp-11 , By : Carlos Galdino
Date : March 29 2020, 07:55 AM
I hope this helps . In C++11, let's say I have a function that takes a functor/function and it forwards the arguments to that function. I can pass in a pointer to std::make_shared, which causes it to instantiate a class and call its contructor. But what is the equivalent to doing this with a raw operator new instead? It seems like std::allocator is what I should use, but the construct function there requires a pointer to the allocated memory. I can make it work by wrapping the new call into the template function new_object, but it seems like there should be a way for me to retrieve that function pointer directly. , Why not something like this?
   std::allocator<foo> alloc;
   do_something(std::bind(&std::allocator<foo>::construct<foo, int, std::string>, &alloc, alloc.allocate(sizeof(foo)), std::placeholders::_1, std::placeholders::_2), 12, "test2");

boost function to any functor with template operator ()

Tag : cpp , By : user183954
Date : March 29 2020, 07:55 AM
I hope this helps you . boost::function, just like std::function requires a specific functor signature (in your case void(void*)), which is the signature it will try to call your functor with. That's why T is deduced as void and the compiler refuses to give you a void&, rightfully so.
Your example is fundamentally add odds with itself, because if you want to have type erasure, you cannot have templates, as the compiler cannot know which templates to instantiate otherwise. Assume for a moment that boost::function could do what you want.
void foo(boost::function<magic> func)
    double a = 5;   

Can I pass a Functor class(that overloads operator()) to a function, that wants a function pointer? And how

Tag : cpp , By : user185939
Date : March 29 2020, 07:55 AM
hop of those help? Traditional way for callback to allow extra parameter is to have extra void* that user provides.
using callback = bool (int, int, void* userData);

bool run_callback(int a, int b, void* userData) {
    CallBack* c = reinterpret_cast<void*>(userData);

    return (*c)(a, b);

CallBack t(true);
Register(&run_callback, &t); // Would call later run_callback(2, 3, &t);
Callback* global = nullptr;

bool RunCallBack(int foo, int bar)
    assert(global != nullptr);
    return (*global)(foo, bar);
Callback f(false);
Callback t(true);

global = &f;
print_result(&RunCallBack, 2, 3);
global = &t;
print_result(&RunCallBack, 2, 3);
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