void(*)(void) is a function pointer while I suppose void(void) is also a way to represent function type. It is used as template parameter in std::function <functional>

What is void(void) and how it is different from void(*)(void)?

CodePudding user response：

Let's start by saying, that sometimes the void(void) will be automatically treated as a pointer to function (i.e. in a parameter list). Then we still can be explicit and use the void(*)(void), but the void(void) will be an equivalent.

// Here, we explicitly state that the f is a pointer to function
void foo(void (*f)(void), int i); 
// Equivalent, f is automatically treated as a pointer to the function
void foo(void f(void), int i);

Which is not the case for the std::function instantiation. The types of the two are different, yet may have similar behavior (i.e. we can use the function call operator on a pointer to function).

void bar();

// Prints 1
std::cout << std::is_same<void(void), decltype(bar)>::value << '\n'; 
// Prints 1 as well
// Note the pointer type specifier
std::cout << std::is_same<void(*)(void), decltype(bar)*>::value << '\n';
// Prints 0
// Note the absence of the pointer type specifier
std::cout << std::is_same<void(*)(void), decltype(bar)>::value << '\n';

The use of the std::function instead of the function pointers has the same moto as using the smart pointers instead of the raw pointers.

N.B. unlike in the case of a parameter, the compiler won't treat the function type in a return as a pointer to function type. Thus, we must explicitly specify the pointer type in case of the return.

// Here, baz is a function that doesn't take any agruments
// And returns a pointer to a function that takes an int argument and "returns" void
void (*baz())(int);

CodePudding user response：

for void fun(){}, std::is_same_v<void(void)>, decltype(fun)> is true; and std::is_same_v<void(*)(void)>, decltype(&fun)> is true. In fact, those are their real types. However, the standard approve you convert an object that has type void(void) to void(*)(void) implicitly (so you can even write (******funptr)()), that's why we always confuse them.

CodePudding user response：

What is void(void) and how it is different from void(*)(void)?

They are distinct types. Although the former(void(void)) can be implicitly converted to the latter(void(*)(void)) in many contexts.

The type void(void) is called a function type. The following are also function types:

int (int, int) //this is a function type
void(int)      //this is a function type as well

On the other hand, void (*)(void) is a pointer to a function type. That is, it is a pointer that points to a function with 0 parameters and return type of void.