function pointer with generic argument type

Question

for knowledge sake, I would like to know if something like this is possible:

2 function:

static int func1(int *a, int b){
...
}

static int func2(double *a, int b){
...
}

I would like to declare a function pointer and point it to one of these function. However the arguments of these functions are of different type. So I tried:

static int (*func_ptr)(void *arg1, int arg2);
static void *Argument1;
static int Argument2=5;
int main(){
 double arg1_d;;
 int arg1_i;
 ...
 if(want_func2){
  Argument1=(double *) &arg1_d;
  func_ptr=func2;
  run(func_ptr);
 }
 else{
  Argument1=(int *) &arg1_i;
  func_ptr=func1;
  run(func_ptr);
 }
 return 0;
}

static int run(int *(function)(void *,int )){
function(Argument1,Argument2);
}

However, when compiling I get the warning:

warning: assignment from incompatible pointer type

when

func_ptr=func2;
func_ptr=func1;

Is there anyway to create a function pointer with a generic type argument?

Maybe run(int *(function)())?

chux
– chux

2014-12-11 16:50:24 +00:00
Commented Dec 11, 2014 at 16:50 — chux
– chux, Commented Dec 11, 2014 at 16:50

Paul92 · Accepted Answer · 2014-12-11 16:22:41Z

2

You are on the right track. But, in order to achieve what you want, your functions should have a void* parameter and cast it internally to int or double.

answered Dec 11, 2014 at 16:22

Paul92

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Comments

aschepler · Accepted Answer · 2014-12-11 16:20:49Z

-1

No. You need the if...else to use such a thing correctly anyway, so just store separate function pointers and use the correct pointer in each piece of code.

answered Dec 11, 2014 at 16:20

aschepler

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1 Comment

Governa Over a year ago

There are valid cases where this is wanted... think the first parameter as an "self" parameter for example.

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