Typedef function pointer?

typedef is a language construct that associates a name to a type.
You use it the same way you would use the original type, for instance

typedef int myinteger;
typedef char *mystring;
typedef void (*myfunc)();

using them like

myinteger i;   // is equivalent to    int i;
mystring s;    // is the same as      char *s;
myfunc f;      // compile equally as  void (*f)();

As you can see, you could just replace the typedefed name with its definition given above.

The difficulty lies in the pointer to functions syntax and readability in C and C++, and the typedef can improve the readability of such declarations. However, the syntax is appropriate, since functions - unlike other simpler types - may have a return value and parameters, thus the sometimes lengthy and complex declaration of a pointer to function.

The readability may start to be really tricky with pointers to functions arrays, and some other even more indirect flavors.

To answer your three questions

• Why is typedef used? To ease the reading of the code - especially for pointers to functions, or structure names.

• The syntax looks odd (in the pointer to function declaration) That syntax is not obvious to read, at least when beginning. Using a typedef declaration instead eases the reading

• Is a function pointer created to store the memory address of a function? Yes, a function pointer stores the address of a function. This has nothing to do with the typedef construct which only ease the writing/reading of a program ; the compiler just expands the typedef definition before compiling the actual code.

Example:

typedef int (*t_somefunc)(int,int);

int product(int u, int v) {
  return u*v;
}

t_somefunc afunc = &product;
...
int x2 = (*afunc)(123, 456); // call product() to calculate 123*456

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"