int
main(int argc, char **argv)
{
printf("sizeof ary=%d\n", sizeof(ary));
printf("sizeof foo=%d\n", sizeof(struct foo));
//fun2(fee);
//(int *)(fee.data[0]) = 55;
//(char **)(fee.data[0]) = "hello";
//(char **)(fee.data[sizeof(int)]) = "hello";
//(char *)(fee.data[sizeof(int) + sizeof(char *)]) = 'k';
//(double *)(fee.data[sizeof(int) + sizeof(char *) + sizeof(double)]) = 178.213;
//((foo_fun)function)(fee);
bee.arg1 = 55;
bee.arg2 = "hello";
bee.arg3 = 'k';
bee.arg4 = 178.213;
((bar_fun)function)(bee);
{
int offset = 0;
poke_int(&fee, &offset, 66);
poke_ptr(&fee, &offset, "goodbye");
poke_char(&fee, &offset, 'j');
poke_double(&fee, &offset, 91.3171);
}
((foo_fun)function)(fee);
}
int main()
{
bank *b = malloc(sizeof(bank));
init_bank(b,100);
poke_int(b,97);
poke_int(b,100);
printf("%d\n",(int)*b->data);
printf("%d\n",peek_int(b,1));
poke_string(b,"Hello World!");
printf("%s\n",&(b->data[4]));
delete_bank(b);
/* I need to learn why when passing bank *b; to a function, then inside the function calling malloc at
* that pointer, the memory wasn't being allocated
*
* Aha! I should of known the answer to this.. Obvious we get a copy of the variables passed to the function
* and the case is the same for pointers. So the local copy gets the allocated memory and the original doesnt
* know about it.
*
* There are ways around it, pass in a pointer to a pointer I want to allocate
*
* void init(bank **b){
* *b = (bank*) malloc(sizeof(bank));
* (*b)->xxx = 0;
* }
*
* or obviously allocate a tmp variable inside and return that.
*
* */
return 0;
}