How to interpret “void(*)()”?

Question

When I read the shared_ptr , I found a piece of code:

void(*)()

How to interpret it?

Answer 1

It is a pointer to a function type, which can be used for all the functions which have no arguments and returns void .

For example:

void function_1() {}
void function_2() {}

void(*func_1_ptr)() = function_1; // or = &function_1;
void(*func_2_ptr)() = function_2; // or = &function_2;

Now func_1_ptr holds the pointer to the function function_1 , and func_2_ptr holds the pointer to function_2 .

You can make the type more intuitive by using declaration.

using FunPtrType = void(*)();

and now you could write

FunPtrType  func_1_ptr = function_1; // or = &function_1;
//Type      identifier   function
FunPtrType  func_2_ptr = function_2; // or = &function_2;

Answer 2

This is the type of a pointer to a function, which takes no arguments and returns void .

Answer 3

The asterisk in between an open-close parenthesis (*) represents the declaration of a function-pointer. The left and right of this represent the return type and function arguments of the function that it will point to.

So basically in your case:

void printHello()
{
    std::cout<<"Hello";
}

void(*fPtr)() = printHello;

In C++ you can do the same using a better OOP way:

std::function<void()> fPtr = printHello;
fPtr();

You will have to include the functional header