Typedef function pointer?

typedef function pointer c++11
typedef function pointer in c - geeksforgeeks
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what is the use of 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?

C Language, Thus the typedef allows a simpler syntax when dealing with function pointers. That's a function that takes two arguments — an int and a pointer to a function which takes an int as an argument and returns nothing — and which returns a pointer to function like its second argument. After the typedef, you can just do this to have 5 function pointers with the same signature: print_function_ptr p1, p2, p3, p4, p5; – blubberdiblub Sep 24 '16 at 17:39 add a comment | 3

  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"
    

Function Pointers in C, Just as a variable can be declared to be a pointer to an int, a variable can also Notice that it is often convenient to use a typedef to define the function type. c++ function-pointers typedef glm-math. share | improve this question | follow | edited 1 hour ago. Nicol Bolas. 341k 45 45 gold badges 574 574 silver badges 763 763

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.

Typedef Function Type, There's no need to typedef pointers to function types, typedefing a function type makes things clearer. 'function' is a pointer to a function type: Thats what the function pointer typedef is for. The alias is the handler which is a typedef for a void function taking the two arguments. std::sort is another example, though its not restricted to function pointers. Function pointers are rarely used in C++, its mostly C where you use them. C++ has std::function, functors and lambdas.

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)>;

"typedef" should be used for function pointers, Function pointer syntax can be hard on the eyes, particularly when one function is used as a parameter to another. Providing and using a typedef instead (or a  We also discussed calling a function using function pointer, passing a function to another function as an argument using function pointer, typedef of function pointers and array of function pointers. A function pointer or pointer to function in C is a usual pointer variable that points to the address of a function in memory.

#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) );
}

How Do I Declare A Function Pointer in C?, typedef returnType typeName(parameterTypes);. (example code). This site is not intended to be an exhaustive list of all possible uses of function pointers. 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.

The Function Pointer Tutorials - Syntax, The Function Pointer Tutorials: Introduction to C and C++ Function Pointers, using a typedef: Define a pointer to a function which is taking // two floats and  typedef can be used to give an alias name to pointers also. Here we have a case in which use of typedef is beneficial during pointer declaration. In Pointers * binds to the right and not on the left. int* x, y; By this declaration statement, we are actually declaring x as a pointer of type int, whereas y will be declared as a plain int variable.

typedef with functions - C++ Forum, would one use them instead of typedef pointer functions? Jan 8, 2013 at 10:48pm​. In this case, as long as the user calls selectionSort normally (not through a function pointer), the comparisonFcn parameter will default to ascending. Making function pointers prettier with typedef or type aliases. Let’s face it -- the syntax for pointers to functions is ugly.

Function Pointers, Declaring with Typedefs. How to declare without typedefs. Remember when you declare a pointer, you have to specify the type of data object you are pointing  typedef would look like this: type TypeFunc : ^((int, int) -> int); The type operator ^ means "pointer to", while X -> Y means "a function taking X as an argument and returning Y." This is quite a bit closer to

Comments
  • Take a look at the link (last section) learncpp.com/cpp-tutorial/78-function-pointers
  • Should be noted that since c++11 using FunctionFunc = void (*)(); can be used instead. It is a bit more clear that you are just declaring a name for a type (pointer to function)
  • just to add to @user362515, a bit clearer form to me is: using FunctionFunc = void(void);
  • @topspin IIRC these two are not the same. One is a function pointer type, the other is function type. There is implicit conversion, that's why it works, IANA(C++)L so, one can step in and correct me. In any case, if the intend is to define a pointer type I think the syntax with the * is a bit more explicit.
  • in the last example, wouldn't just 'square' refer to the same thing i.e pointer to the function instead of using &square.
  • Question, in your first typedef example you have of the form typedef type alias but with function pointers there only seems to be 2 arguments, typedef type. Is alias defaulted to the name specified in type argument?
  • @pranavk: Yes, square and &square (and, indeed, *square and **square) all refer to the same function pointer.
  • @user814628: It is not clear quite what you're asking. With typedef int newname, you are making newname into an alias for int. With typedef int (*func)(int), you are making func into an alias for int (*)(int) — a pointer to function taking an int argument and returning an int value.
  • I guess I'm just confused about the ordering. With typedef int (*func)(int), I understand that func is an alias, just a little confused because the alias is tangled with the type. Going by typedef int INT as an example I would be more of ease if typedef function pointer was of form typedef int(*function)(int) FUNC_1. That way I can see the type and alias in two separate token instead of being meshed into one.
  • typedef does not declare a new type. you can have many typedef-defined names of the same type, and they are not distinct (e.g. wrt. overloading of functions). there are some circumstances in which, with respect to how you can use the name, a typedef-defined name is not exactly equivalent to what it's defined as, but multiple typedef-defined names for the same, are equivalent.
  • Ah i get it now. Thanks.
  • +1 for the answer to question 2 - very clear. The rest of the answers were clear too, but that one stands out for me :-)