Function pointer to different functions with different arguments in C

Question

I have two functions with variable number and types of arguments

double my_func_one(double x, double a, double b, double c) { return x + a + b + c }
double my         


        

Answer 1

An old topic but I am facing same issue and finally came with this idea.

If you want the same prototype for each functions you can wrap the parameters in structure like this :

 typedef struct {
     double a,
            b,
            c; 
 }chunk1_t;

  typedef struct {
     double p[];
     double c; 
 }chunk2_t;

Then your function pointer becomes:

 double (*pfunc) (double x, void *args);

which leads to something like this :

 pfunc cb;

 double swap_function(double x, pfunc cb, void *args);

 double my_func_one(double x, void *args) { 
   chunk1_t *chunk = (chunk1_t*) args;
   return x + chunk->a + chunk->b + chunk->c; 
 }

 double my_func_two(double x, void *args) {
   chunk2_t *chunk = (chunk2_t*) args;
   return x + chunk->p[0] + chunk->p[1] + chunk->c ;
 }

 int main(){
   // declare a, b,...
   double a = 1.0f;
   //...
   // cast for safety
   chunk1_t myChunk1 = {(double)a, (double)b, (double)c};
   // don't if you like risk
   chunk2_t myChunk2 = {p, c};

   swap_function(x, cb, &myChunk1); 
   swap_function(x, cb, &myChunk2);
 }

Using function pointer stored in structure:

 #define FUNC_ONE_METHOD 1
 #define FUNC_TWO_METHOD 2

 typedef struct chunk{
     double a, b, c;
     double p[];
     int method;
     double (*pfunc) (double x, struct chunk *self);
 }chunk_t;

 double swap_function(double x, chunk_t *ch){
    switch (ch->method)
    {
        case FUNC_TWO_METHOD:
            ch->pfunc = my_func_two;
            break;
        case FUNC_ONE_METHOD:
            ch->pfunc = my_func_one;
            break;
        default:
            return -1; // or throw error
    return ch->pfunc(x, ch);
 }


 chunk c = {.a= 1, .b=3, .c=1, .method=1};
 swap_function(x, &c);

Answer 2

My question is, for this scenario, Can I at all define a function pointer?

No. (Other than by dirty typecasting.)

Is there any other way to do this?

Your best bet is to create a wrapper function for one of your existing functions. For example:

double my_func_one_wrapper(double x, double p[], double c) {
    return my_func_one(x, p[0], p[1], c);
}

That way, you have two functions with the same signature, and therefore the same function-pointer type.

Answer 3

Your understanding is true.
The signature of your function-pointer must match the corresponding function(s).

Consider learncpp.com:

Just like it is possible to declare a non-constant pointer to a variable, it’s also possible to >declare a non-constant pointer to a function. The syntax for doing so is one of the ugliest things >you will ever see:

// pFoo is a pointer to a function that takes no arguments and returns an integer 
int (*pFoo) (); 

The parenthesis around *pFoo are necessary for precedence reasons, as int *pFoo() would be interpreted as a function named pFoo that takes no parameters and returns a pointer to an integer.

In the above snippet, pFoo is a pointer to a function that has no parameters and returns an integer. pFoo can “point” to any function that matches this signature.

...

Note that the signature (parameters and return value) of the function pointer must match the signature of the function.

Answer 4

A sample of Typecasting approach for using a same function pointer for different functions of different prototypes. <>

#include <stdio.h>
typedef void (*myFuncPtrType) (void);

typedef int (*myFunc2PtrType)(int, int);

typedef int * (*myFunc3PtrType)(int *);

static void myFunc_1 (void);
static int myFunc_2 (int, int);
static int* myFunc_3 (int *);

const myFuncPtrType myFuncPtrA[] = {
                                    (myFuncPtrType)myFunc_1,
                                    (myFuncPtrType)myFunc_2,
                                    (myFuncPtrType)myFunc_3
};

static void myFunc_1 (void)
{
    printf("I am in myFunc_1 \n");
}

static int myFunc_2 (int a, int b)
{
    printf("I am in myFunc_2\n");
    return (a+b);
}

static int* myFunc_3 (int *ptr)
{
    printf("I am in myFunc_3\n");
    *ptr = ((*ptr) * 2);
    return (ptr+1);
}

int main(void) {
    // your code goes here
    int A[2],C;

    int* PTR = A;

    (*(myFuncPtrA[0]))();

    A[0]=5;
    A[1]=6;

    C = ((myFunc2PtrType)(*(myFuncPtrA[1])))(A[0],A[1]);

    printf("Value of C: %d \n", C);

    printf("Value of PTR before myFunc_3: %p \n", PTR);
    printf("Value of *PTR before myFunc_3: %d \n", *PTR);

    PTR = ((myFunc3PtrType)(*(myFuncPtrA[2])))(&A);

    //Lets look how PTR has changed after the myFunc_3 call

    printf("Value of PTR after myFunc_3: %p \n", PTR);
    printf("Value of *PTR after myFunc_3: %d \n", *PTR);


    return 0;
}

Answer 5

What you want is possible, but a bit dirty. Function pointers can be cast to one another without losing information. The important thing is that you'd always have to call a function through such a pointer only with a signature that corresponds to its definition. So if you cast back before calling the function(s) and call with the correct arguments, all should be fine.

Answer 6

The cleanest way to do it is to use a union:

typedef union {
  double (*func_one)(double x, double a, double b, double c);
  double (*func_two)(double x, double p[], double c);
} func_one_two;

Then you can initialize an instance of the union, and include information to the swap_function function to say which field is valid:

func_one_two func;

if (condition_1)
   func.func_one = my_func_one;
else if (condition_2)
   func.func_two = my_func_two;

 // The function that will use the function I passed to it
 swap_function(a, b, func, condition_1);

This assumes that swap_function can know based on condition_1 being false that it should assume condition_2. Note that the union is passed by value; it's only a function pointer in size after all so that's not more expensive than passing a pointer to it.