difference between void(int) & void (*)(int)

Question

I know void (*)(int) is to function pointer but what is void(int) ?

It's used for std::function template.

Say I have a function void fun(int){} : decltype(&fun) gives void(*)(int) but decltype(fun) gives void(int)

Answer 1

If T is a type, then T* denotes the type "pointer-to- T ".

The type void(int) is a function type , it's the type of a function taking one int and returning void . For example, it is the type of f if f is declared as void f(int);

If T = void(int) , then T* is spelled void(*)(int) , so the latter is the type of a function pointer . You can also form a reference to a function, which is T& = void(&)(int) ; this is occasionally more useful (eg you can take the address of a function lvalue).

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Aside note: Function lvalues decay to their function pointer very easily. You can call a function either via a function lvalue or via a function pointer. When used as an operand for the indirection operator ( * ), the function value decays, so you can dereference the pointer again and again:

printf("Hello world\n");        // OK
(*printf)("Hello world\n");     // also OK
(****printf)("Hello world\n");  // four-star programmer

Some of the only times that a function does not decay is when used as the operand of the address-of operator, or when bound to a reference:

void f(int);          // our example function

void(*p1)(int) = &f;  // no decay of "f" here
void(*p2)(int) = f;   // "f" decays
void(&r1)(int) = f;   // no decay of "f" here

void g(void(&callback)(int), int n) {
  callback(n);
}
g(f, 10);             // no decay of "f" here

template <typename F, typename ...Args>
decltype(auto) h(F&& callback, Args&&... args) {
    return std::forward<F>(callback)(std::forward<Args>(args)...);
}
h(f, 10);             // no decay of "f" here

Answer 2

void (*whatever)(int) 

should be read as: whatever is a pointer, pointing to a function, that accepts one int as argument, and returns nothing (ie., void).

void whatever(int)

should be read as: whatever is a function (NOT a pointer), that accepts one int as argument, and returns nothing (ie., void)

Once the pointer to a function is initialized to point to a valid function (one that satisfies the prototype), then you can invoke the function either through its "real" name, or through the pointer.

Pointers to functions are very useful - they're variables, just like anything else, so you can pass them around to other functions (see eg qsort()), you can put them in structs, etc..

Given this, the following code is valid:

#include <stdio.h>

void myfun(int x) {
  printf("The value of X is %d\n", x);
}

int main() {
    void (*myfunp)(int);

    myfunp = &myfun;

    myfun(13);
    myfunp(12);

    return 0;
}

Answer 3

void(*)(int) should be read as type of a pointer which is pointing to a function , that accepts one int as argument, and returns nothing.

For understanding more on function to pointer and its usage please check here: http://www.cprogramming.com/tutorial/function-pointers.html