[c++] Typedef function pointer?

I'm learning how to dynamically load DLL's but what I don't understand is this line

typedef void (*FunctionFunc)();

I have a few questions. If someone is able to answer them I would be grateful.

1. Why is typedef used?
2. The syntax looks odd; after void should there not be a function name or something? It looks like an anonymous function.
3. Is a function pointer created to store the memory address of a function?

So I'm confused at the moment; can you clarify things for me?

#include <stdio.h>
#include <math.h>

/*
To define a new type name with typedef, follow these steps:
1. Write the statement as if a variable of the desired type were being declared.
2. Where the name of the declared variable would normally appear, substitute the new type name.
3. In front of everything, place the keyword typedef.
*/

// typedef a primitive data type
typedef double distance;

// typedef struct 
typedef struct{
    int x;
    int y;
} point;

//typedef an array 
typedef point points[100]; 

points ps = {0}; // ps is an array of 100 point 

// typedef a function
typedef distance (*distanceFun_p)(point,point) ; // TYPE_DEF distanceFun_p TO BE int (*distanceFun_p)(point,point)

// prototype a function     
distance findDistance(point, point);

int main(int argc, char const *argv[])
{
    // delcare a function pointer 
    distanceFun_p func_p;

    // initialize the function pointer with a function address
    func_p = findDistance;

    // initialize two point variables 
    point p1 = {0,0} , p2 = {1,1};

    // call the function through the pointer
    distance d = func_p(p1,p2);

    printf("the distance is %f\n", d );

    return 0;
}

distance findDistance(point p1, point p2)
{
distance xdiff =  p1.x - p2.x;
distance ydiff =  p1.y - p2.y;

return sqrt( (xdiff * xdiff) + (ydiff * ydiff) );
}

1. typedef is used to alias types; in this case you're aliasing FunctionFunc to void(*)().

2. Indeed the syntax does look odd, have a look at this:

   typedef   void      (*FunctionFunc)  ( );
   //         ^                ^         ^
   //     return type      type name  arguments
   

3. No, this simply tells the compiler that the FunctionFunc type will be a function pointer, it doesn't define one, like this:

   FunctionFunc x;
   void doSomething() { printf("Hello there\n"); }
   x = &doSomething;
   
   x(); //prints "Hello there"
   

Without the typedef word, in C++ the declaration would declare a variable FunctionFunc of type pointer to function of no arguments, returning void.

With the typedef it instead defines FunctionFunc as a name for that type.

For general case of syntax you can look at annex A of the ANSI C standard.

In the Backus-Naur form from there, you can see that typedef has the type storage-class-specifier.

In the type declaration-specifiers you can see that you can mix many specifier types, the order of which does not matter.

For example, it is correct to say,

long typedef long a;

to define the type a as an alias for long long. So , to understand the typedef on the exhaustive use you need to consult some backus-naur form that defines the syntax (there are many correct grammars for ANSI C, not only that of ISO).

When you use typedef to define an alias for a function type you need to put the alias in the same place where you put the identifier of the function. In your case you define the type FunctionFunc as an alias for a pointer to function whose type checking is disabled at call and returning nothing.

If you can use C++11 you may want to use std::function and using keyword.

using FunctionFunc = std::function<void(int arg1, std::string arg2)>;