Another void* topic; I just have to ask because I am confused

Question

Ok, muddling though Stack on the particulars about void*, books like The C Programming Language (K&R) and The C++ Programming Language (Stroustrup). What have I learned? That void* is a generic pointer with no type inferred. It requires a cast to any defined type and printing void* just yields the address.

What else do I know? void* can't be dereferenced and thus far remains the one item in C/C++ from which I have discovered much written about but little understanding imparted.

I understand that it must be cast such as *(char*)void* but what makes no sense to me for a generic pointer is that I must somehow already know what type I need in order to grab a value. I'm a Java programmer; I understand generic types but this is something I struggle with.

So I wrote some code

typedef struct node
{
  void* data;
  node* link;
}Node;

typedef struct list
{
   Node* head;
}List;

Node* add_new(void* data, Node* link);

void show(Node* head);

Node* add_new(void* data, Node* link)
{
  Node* newNode = new Node();
  newNode->data = data;
  newNode->link = link;

  return newNode;
}

void show(Node* head)
{
  while (head != nullptr)
  {
      std::cout << head->data;
      head = head->link;
  }
}

int main()
{
  List list;

  list.head = nullptr;

  list.head = add_new("My Name", list.head);

  list.head = add_new("Your Name", list.head);

  list.head = add_new("Our Name", list.head);

  show(list.head);

  fgetc(stdin);

  return 0;
}

I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out? This implies I already need to know the type, and this reveals nothing about the generic nature of void* while I follow what is here although still no understanding.

Why am I expecting void* to cooperate and the compiler to automatically cast out the type that is hidden internally in some register on the heap or stack?

I think the use of `void*` is: Window's CWnd class can hold a `void* data`. It doesn't know what this data is, nor does it care. It doesn't do anything with this `data`, it merely holds it for me. Meanwhile, my code can store my `thingamabob` into this `data`. My code assumes that the `data` always holds a `thingamabob`, which is fine because my code is the only code accessing this `data` member. My code "knows" the type, but `CWnd` doesn't. — Mooing Duck, Sep 27 '17 at 17:45

R Sahu · Accepted Answer · 2017-09-27T19:43:29.000

5

I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out?

You can't. You must know the valid types that the pointer can be cast to before you can dereference it.

Here are couple of options for using a generic type:

If you are able to use a C++17 compiler, you may use std::any.
If you are able to use the boost libraries, you may use boost::any.

edited Sep 27 '17 at 19:43

answered Sep 27 '17 at 17:26

R Sahu

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Is there no way to practice `pure` generics in c or c++? – Mushy Sep 27 '17 at 17:30
2

@Mushy, no , not in the core language. There are attempts to make generic object types in the forms of [`std::any`](http://en.cppreference.com/w/cpp/utility/any) and [`boost::any`](http://www.boost.org/doc/libs/1_61_0/doc/html/any.html) – R Sahu Sep 27 '17 at 17:34
1

@RSahu there is `std::any` in C++17 – Swift - Friday Pie Sep 27 '17 at 19:07

Fabien Bouleau · Answer 2 · 2017-09-27T17:43:45.330

Unlike Java, you are working with memory pointers in C/C++. There is no encapsulation whatsoever. The void * type means the variable is an address in memory. Anything can be stored there. With a type like int * you tell the compiler what you are referring to. Besides the compiler knows the size of the type (say 4 bytes for int) and the address will be a multiple of 4 in that case (granularity/memory alignment). On top, if you give the compiler the type it will perform consistency checks at compilation time. Not after. This is not happening with void *.

In a nutshell, you are working bare metal. The types are compiler directives and do not hold runtime information. Nor does it track the objects you are dynamically creating. It is merely a segment in memory that is allocated where you can eventually store anything.

score 0 · Answer 3 · answered Sep 27 '17 at 17:34

The main reason to use void* is that different things may be pointed at. Thus, I may pass in an int* or Node* or anything else. But unless you know either the type or the length, you can't do anything with it.

But if you know the length, you can handle the memory pointed at without knowing the type. Casting it as a char* is used because it is a single byte, so if I have a void* and a number of bytes, I can copy the memory somewhere else, or zero it out.

Additionally, if it is a pointer to a class, but you don't know if it is a parent or inherited class, you may be able to assume one and find out a flag inside the data which tells you which one. But no matter what, when you want to do much beyond passing it to another function, you need to cast it as something. char* is just the easiest single byte value to use.

Swift - Friday Pie · Answer 4 · 2017-09-27T19:10:50.743

Your confusion derived from habit to deal with Java programs. Java code is set of instruction for a virtual machine, where function of RAM is given to a sort of database, which stores name, type, size and data of each object. Programming language you're learning now is meant to be compiled into instruction for CPU, with same organization of memory as underlying OS have. Existing model used by C and C++ languages is some abstract built on top of most of popular OSes in way that code would work effectively after being compiled for that platform and OS. Naturally that organization doesn't involve string data about type, except for famous RTTI in C++.

For your case RTTI cannot be used directly, unless you would create a wrapper around your naked pointer, which would store the data.

In fact C++ library contains a vast collection of container class templates that are useable and portable, if they are defined by ISO standard. 3/4 of standard is just description of library often referred as STL. Use of them is preferable over working with naked pointers, unless you mean to create own container for some reason. For particular task only C++17 standard offered std::any class, previously present in boost library. Naturally, it is possible to reimplement it, or, in some cases, to replace by std::variant.

score 0 · Answer 5 · answered Sep 27 '17 at 20:19

Assuming I have no understanding of the type stored in void*, how do I get the value out

You don't.

What you can do is record the type stored in the void*.

In c, void* is used to pass around a binary chunk of data that points at something through one layer of abstraction, and recieve it at the other end, casting it back to the type that the code knows it will be passed.

void do_callback( void(*pfun)(void*), void* pdata ) {
  pfun(pdata);
}

void print_int( void* pint ) {
  printf( "%d", *(int*)pint );
}

int main() {
  int x = 7;
  do_callback( print_int, &x );
}

here, we forget thet ype of &x, pass it through do_callback.

It is later passed to code inside do_callback or elsewhere that knows that the void* is actually an int*. So it casts it back and uses it as an int.

The void* and the consumer void(*)(void*) are coupled. The above code is "provably correct", but the proof does not lie in the type system; instead, it depends on the fact we only use that void* in a context that knows it is an int*.

In C++ you can use void* similarly. But you can also get fancy.

Suppose you want a pointer to anything printable. Something is printable if it can be << to a std::ostream.

struct printable {
  void const* ptr = 0;
  void(*print_f)(std::ostream&, void const*) = 0;

  printable() {}
  printable(printable&&)=default;
  printable(printable const&)=default;
  printable& operator=(printable&&)=default;
  printable& operator=(printable const&)=default;

  template<class T,std::size_t N>
  printable( T(&t)[N] ):
    ptr( t ),
    print_f( []( std::ostream& os, void const* pt) {
      T* ptr = (T*)pt;
      for (std::size_t i = 0; i < N; ++i)
        os << ptr[i];
    })
  {}
  template<std::size_t N>
  printable( char(&t)[N] ):
    ptr( t ),
    print_f( []( std::ostream& os, void const* pt) {
      os << (char const*)pt;
    })
  {}
  template<class T,
    std::enable_if_t<!std::is_same<std::decay_t<T>, printable>{}, int> =0
  >
  printable( T&& t ):
    ptr( std::addressof(t) ),
    print_f( []( std::ostream& os, void const* pt) {
      os << *(std::remove_reference_t<T>*)pt;
    })
  {}
  friend
  std::ostream& operator<<( std::ostream& os, printable self ) {
    self.print_f( os, self.ptr );
    return os;
  }
  explicit operator bool()const{ return print_f; }
};

what I just did is a technique called "type erasure" in C++ (vaguely similar to Java type erasure).

void send_to_log( printable p ) {
  std::cerr << p;
}

Live example.

Here we created an ad-hoc "virtual" interface to the concept of printing on a type.

The type need not support any actual interface (no binary layout requirements), it just has to support a certain syntax.

We create our own virtual dispatch table system for an arbitrary type.

This is used in the C++ standard library. In c++11 there is std::function<Signature>, and in c++17 there is std::any.

std::any is void* that knows how to destroy and copy its contents, and if you know the type you can cast it back to the original type. You can also query it and ask it if it a specific type.

Mixing std::any with the above type-erasure techinque lets you create regular types (that behave like values, not references) with arbitrary duck-typed interfaces.

Another void* topic; I just have to ask because I am confused

5 Answers5