Typedef function pointer? - c++

Question

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.
Why is typedef used?
The syntax looks odd; after void should there not be a function name or something? It looks like an anonymous function.
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?

Answer 1

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

Answer 2

typedef is used to alias types; in this case you're aliasing FunctionFunc to void(*)().
Indeed the syntax does look odd, have a look at this:
typedef void (*FunctionFunc) ( );
// ^ ^ ^
// return type type name arguments
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"

Answer 3

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.

Answer 4

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)>;

Answer 5

#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) );
}

Answer 6

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.