fill an array of pointers to functions in another function in C

Question

I tried to fill my array of pointer to functions in another function but I don't know how I can do it.

Below my function to fill the array

void* fill_tab_f(void (***tab_f))
{

    *tab_f[0] = ft_pt_char;
    *tab_f[1] = ft_pt_str;
    *tab_f[2] = ft_pt_ptr;
    *tab_f[3] = ft_pt_int;
    *tab_f[4] = ft_pt_int;
    *tab_f[5] = ft_pt_un_int;
    *tab_f[6] = ft_pt_hexa_min;
    *tab_f[7] = ft_pt_hexa_maj;
    
    return NULL;
}

Below the declaration of the array of pointers to functions and the call of the function to fill my array.

void(*tab_f[8])(va_list *, Variable*);
fill_tab_f(&tab_f);

Thanks for your answers.

Answer 1

You are overcomplicating it

typedef int variable ;

void fill_tab_f(void (*tab_f[])(va_list *, variable *))
{

    tab_f[0] = ft_pt_char;
    tab_f[1] = ft_pt_str;
    tab_f[2] = ft_pt_ptr;
    tab_f[3] = ft_pt_int;
    tab_f[4] = ft_pt_int;
    tab_f[5] = ft_pt_un_int;
    tab_f[6] = ft_pt_hexa_min;
    tab_f[7] = ft_pt_hexa_maj;
}

or if function pointer syntax is a bit weird for you:

typedef int variable ;

typedef void (*funcptr)(va_list *, variable *);

void fill_tab_f(funcptr *tab_f)
{

    tab_f[0] = ft_pt_char;
    tab_f[1] = ft_pt_str;
    tab_f[2] = ft_pt_ptr;
    tab_f[3] = ft_pt_int;
    tab_f[4] = ft_pt_int;
    tab_f[5] = ft_pt_un_int;
    tab_f[6] = ft_pt_hexa_min;
    tab_f[7] = ft_pt_hexa_maj;
}